IN  MEMORIAM 
FLORIAN  CAJORI 


ELEMENTS  OF  GRAPHIC  STATICS. 


THE 


ELEMENTS  OF  GRAPHIC  STATICS 


21  STe.xt=33oofc  for  Students  of  Engineering 


BY 


L.    M.    HOSKINS 

PROFESSOR  OF  PURE  AND  APPLIED  MECHANICS  IN  THE  LELAND 

STANFORD  JUNIOR  UNIVERSITY;    FORMERLY  PROFESSOR 

OF  MECHANICS  IN  THE  UNIVERSITY  OF  WISCONSIN 


Xeto  gork 
MACMILLAN    AND    CO. 

AND     LONDON 
1892 

A  U  rights  reserved 


COPYRIGHT,  1892, 
BY  MACMILLAN  AND  CO. 


ORI 


TYPOGRAPHY, BY  J.  S.  GUSHING  &  .Co.,  BOSTON,  U.S.A. 
PRESSWORK  BY  BERWICK  &  SMITH,  BOSTON,  U.S.A. 


PREFACE. 


THE  present  work  is  designed  as  an  elementary  text-book  for  the 
use  of  students  of  engineering.  In  preparing  it,  a  chief  aim  has  been 
simplicity  of  presentation.  The  matter  treated  has  been  limited  to 
the  development  of  fundamental  principles,  and  their  application  to  the 
solution  of  typical  problems.  The  method  of  the  force  and  funicular 
polygons  is  deduced  purely  from  statical  principles,  with  very  little 
consideration  of  the  geometrical  theory  of  reciprocal  figures.  Since 
the  book  is  designed  to  embrace  only  what  can  profitably  be  taken  in 
an  elementary  course  by  the  student  of  engineering,  it  has  not  been 
thought  best  to  include  a  discussion  of  problems  involving  the  theory 
of  elasticity.  For  similar  reasons,  the  discussion  of  curves  of  inertia 
has  been  limited  to  simple  cases ;  a  more  general  treatment  being  of 
interest  to  few  besides  the  student  of  pure  mathematics.  No  effort 
has  been  made  to  secure  novelty  in  the  matter  treated,  but  it  is 
believed  that  in  a  few  cases  it  has  been  found  possible  to  simplify, 
and  perhaps  thereby  improve,  the  methods  usually  adopted. 

Attention  is  invited  to  the  method  adopted  for  lettering  corre- 
sponding lines  in  force ,  and  space  diagrams.  It  will  be  seen  that  this 
is  merely  an  extension  of  Bow's  well-known  notation.  This  notation 
is,  however,  capable  of  a  much  wider  use  than  has  usually  been  given 
it.  It  is  believed  that  its  use,  wherever  applicable,  will  be  found  of 
great  value,  both  in  facilitating  the  work  of  the  student,  and  in 
guarding  the  draughtsman  against  mistakes. 

There  is  an  unfortunate  diversity  of  usage  among  writers  in  regard 
to  the  technical  terms  of  mechanics,  —  a  diversity  especially  notice- 
able in  engineering  literature.  In  this  book  the  endeavor  has  been 
made  in  all  cases  to  comply  with  the  usage  to  which  the  highest 

authorities  are  tending. 

L.  M.  H. 

MADISON,  Wis.,  July,  1892. 


918236 


CONTENTS. 


PART   I.  — GENERAL   THEORY. 
CHAPTER   I.     DEFINITIONS. — CONCURRENT  FORCES. 


PAGE 


§  i.    Preliminary  Definitions i 

2.  Composition  of  Concurrent  Forces        .     .    ..    ' 5 

3.  Equilibrium  of  Concurrent  Forces 6 

4.  Resolution  of  Concurrent  Forces 8 

CHAPTER  II      NON-CONCURRENT  FORCES. 

§  i.    Composition  of  Ncn-concurrent  Forces  Acting  on  the  Same  Rigid 

Body ii 

2.  Equilibrium  of  Non-concurrent  Forces 18 

3.  Resolution  into  Non-concurrent  Systems 28 

4.  Moments  of  Forces  and  of  Couples 30 

5.  Graphic  Determination  of  Moments 35 

6.  Summary  of  Conditions  of  Equilibrium 37 

CHAPTER  III.     INTERNAL  FORCES  AND  STRESSES. 

§  i.    External  and  Internal  Forces 4° 

2.  External  and  Internal  Stresses 42 

3.  Determination  of  Internal  Stresses  .     .     .     .     ~ 46 

PART   II.  — STRESSES   IN   SIMPLE   STRUCTURES. 

CHAPTER  IV.     INTRODUCTORY. 

§  i .    Outline  of  Principles  and  Methods 53 


PAGE 


viii  CONTENTS. 

CHAPTER  V.     ROOF  TRUSSES.  —  FRAMED  STRUCTURES  SUSTAINING 
STATIONARY  LOADS. 

§  i.    Loads  on  Roof  Trusses 58 

2.  Roof  Truss  with  Vertical  Loads 61 

3.  Stresses  Due  to  Wind  Pressure 67 

4.  Maximum  Stresses 71 

5.  Cases  Apparently  Indeterminate 74 

6.  Three-hinged  Arch 80 

7.  Counterbracing 88 

CHAPTER  VI.     SIMPLE  BEAMS. 

§  i.    General  Principles 94 

2.  Beam  Sustaining  Fixed  Loads 98 

3.  Beam  Sustaining  Moving  Loads 101 


CHAPTER  VII.     TRUSSES  SUSTAINING  MOVING  LOADS. 

§  i .    Bridge  Loads 112 

2.  Truss  Regarded  as  a  Beam 115 

3.  Truss  with  Parallel  Chords  Sustaining  Concentrated  Loads       .     .  117 

4.  Parallel  Chords  —  Uniformly  Distributed  Moving  Load  .     .     .     .  130 

5.  Truss  with  Curved  Chords  —  Uniform  Panel  Loads 132 

6.  Truss  with  Curved  Chords— Concentrated  Loads 137 

PART   III.—  CENTROIDS   AND    MOMENTS   OF   INERTIA. 

CHAPTER  VIII.     CENTROIDS. 

§  i.    Centroid  of  Parallel  Forces 144 

2.  Center  of  Gravity  —  Definitions  and  General  Principles  .     .     .     .  149 

3.  Centroids  of  Lines  and  of  Areas 152 

CHAPTER  IX.     MOMENTS  OF  INERTIA. 

§  i.    Moments  of  Inertia  of  Forces 159 

2.  Moments  of  Inertia  of  Plane  Areas  .    * 169 

CHAPTER  X.     CURVES  OF  INERTIA. 

§  i.    General  Principles 179 

2.  Inertia-Ellipses  for  Systems  of  Forces 182 

3.  Inertia-Curves  for  Plane  Areas 187 


GRAPHIC    STATICS. 

PART    I. 
GENERAL    THEORY. 


CHAPTER   L  — DEFINITIONS.     CONCURRENT 
FORCES. 

§  I.    Preliminary  Definitions. 

1.  Dynamics  treats  of  the  action  of  forces  upon  bodies.     Its 
two  main  branches  are  Statics  and  Kinetics. 

Statics  treats  of  the  action  of  forces  under  such  conditions 
that  no  change  of  motion  is  produced  in  the  bodies  acted  upon. 

Kinetics  treats  of  the  laws  governing  the  production  of  motion 
by  forces. 

2.  Graphic  Statics  has  for  its   object  the   deduction   of   the 
principles  of  statics,  and  the  solution  of  its  problems,  by  means 
of  geometrical  figures. 

3.  A  Force  is  that  which  tends  to  change  the  state  of  motion 
of  a  body.     We  conceive  of  a  force  as  a  push  or  a  pull  applied 
to  a  body  at  a  definite  point  and  in  a  definite  direction.     Such 
a  push  or  pull  tends  to  give  motion  to  the  body,  but  this  ten- 
dency may  be  neutralized  by  the  action  of  other  forces.     The 
effect  of  a  force  is  completely  determined  when  three  things 
are  given,  —  its  magnitude,  its  direction,  and  its  point  of  applica- 
tion.    The  line  parallel  to  the  direction  of  the  force  and  con- 
taining its  point  of  application,  is  called  its  line  of  action. 


:2'.*   ::.*?.*•*  •»•          GRAPHIC  STATICS. 

Every  force  acting  upon  a  body  is  exerted  by  some  other  body. 
But  the  problems  of  statics  usually  concern  only  the  body  acted 
upon.  Hence,  frequently,  no  reference  is  made  to  the  bodies 
exerting  the  forces. 

4.  Unit  Force.  —  The  unit  force  is  a  force  of  arbitrarily  chosen 
magnitude,  in  terms   of  which  forces  are  expressed.     Several 
different  units  are  in  use.     The  one  employed  in  this  work  is 
\.\\Q  pound,  which  will  now  be  denned. 

A  pound  force  is  a  force  equal  to  the  weight  of  a  pound  mass 
at  the  earth's  surface.  A  pound  mass  is  the  quantity  of  matter 
contained  in  a  certain  piece  of  platinum,  arbitrarily  chosen,  and 
established  as  the  standard  by  act  of  the  British  Parliament. 

The  pound  force,  as  thus  denned,  is  not  perfectly  definite, 
since  the  weight  of  any  given  mass  (that  is,  the  attraction  of 
the  earth  upon  it)  is  not  the  same  for  all  positions  on  the  earth's 
surface.  The  variations  are,  however,  unimportant  for  most  of 
the  requirements  of  the  engineer. 

In  its  fundamental  meaning,  the  word  "pound"  refers  to  the 
unit  mass,  and  it  is  unfortunate  that  it  is  also  applied  to  the  unit 
force.  The  usage  is,  however,  so  firmly  established  that  it  will 
be  here  followed. 

5.  Concurrent  and  Non-concurrent  Forces.  —  Forces  acting  on 
the  same  body  are  concurrent  when  they  have  the  same  point  of 
application.     When  applied  at  different  points  they  are  non- 
concurrent. 

6.  Complanar  Forces  are  those  whose  lines  of  action  are  in 
the  same  plane.     In  this  work,  only  complanar  systems  will  be 
considered  unless  otherwise  specified. 

7.  A  Couple  is  the  name  given  to  a  system  consisting  of  two 
forces,  equal  in  magnitude,  but  opposite  in  direction,  and  having 
different  lines  of  action.     The  perpendicular  distance  between 
the  two  lines  of  action  is  called  the  arm  of  the  couple. 


PRELIMINARY   DEFINITIONS.  3 

8.  Equivalent  Systems  of  Forces.  —  Two  systems  of  forces 
are  equivalent  when   either  may  be  substituted  for  the  other 
without  change  of  effect. 

9.  Resultant. — A  single  force  that  is  equivalent  to  a  given 
system  of  forces  is  called  the  resultant  of  that  system.     It  will 
be  shown   subsequently  that  a  system   of  forces  may  not  be 
equivalent  to  any  single  force.     When   such   is  the  case,  the 
simplest  system  equivalent  to  the  given  system  may  be  called 
its    resultant.      Any   forces    having    a    given    force    for   their 
resultant  are  called  components  of  that  force. 

10.  Composition  and  Resolution  of   Forces.  —  Having   given 
any  system  of  forces,  the  process  of  finding  an  equivalent  system 
is  called  the  composition  of  forces,  if  the  system  determined  con- 
tains fewer  forces  than  the  given  system.     If  the  reverse  is  the 
case,  the  process  is  called  the  resolution  of  forces. 

The  process  of  finding  the  resultant  of  any  given  forces  is 
the  most  important  case  of  composition;  while  the  process  of 
finding  two  or  more  forces,  which  together  are  equivalent  to  a 
single  given  force,  is  the  most  common  case  of  resolution. 

11.  Representation  of  Forces  Graphically. — The  magnitude 
and  direction  of  a  force  can  both  be  represented  by  a  line ;  the 
length  of  the  line  representing_the  magnitude  of  the  force,  and 
its  direction  the  direction  of  the  force. 

In  order  that  the  length  of  a  line  may  represent  the  magni- 
tude of  a  force,  a  certain  length  must  be  chosen  to  denote  the 
unit  force.  Then  a  force  of  any  magnitude  will  be  represented 
by  a  length  which  contains  the  assumed  length  as  many  times 
as  the  magnitude  of  the  given  force  contains  that  of  the  unit 
force. 

In  order  that  the  direction  of  a  force  may  be  represented  by 
a  line,  there  must  be  some  means  of  distinguishing  between  the 
two  opposite  directions  along  the  line.  The  usual  method  is  to 
place  an  arrow-head  on  the  line,  pointing  in  the  direction  toward 


4  GRAPHIC   STATICS. 

which  the  force  acts.  If  the  line  is  designated  by  letters  placed 
at  its  extremities,  the  order  in  which  these  are  read  may  indicate 
the  direction  of  the  force.  Thus,  AB  and  BA  represent  two 
forces,  equal  in  magnitude  but  opposite  in  direction. 

The  line  of  action  of  a  force  can  also  be  represented  by  a  line 
drawn  on  the  paper. 

In  solving  problems  in  statics,  it  is  usually  convenient  to 
draw  two  separate  figures,  in  one  of  which  the  forces  are 
represented  in  magnitude  and  direction  only,  and  in  the  other 
in  line  of  action  only. 

These  two  species  of  diagrams  will  be  called  force  diagrams 
and  space  diagrams,  respectively. 

12.  Notation.  —  The  use  of  graphic  methods  is  much  facili- 
tated by  the  adoption  of  a  convenient  system  of  notation  in 
the  figures  drawn. 

There  will  generally  be  two  figures  (the  force  diagram  and 
the  space  diagram)  so  related  that  for  every  line  in  one  there 
is  a  corresponding  line  in  the  other. 

In  the  force  diagram  each  line  represents  a  force  in  magni- 
tude and  direction ;    in  the  space  diagram  the  corresponding 
line  represents   the  line  of  action  of  the  force.     These   lines 
will  usually  be  designated  in  the  following  manner :  The  line 
denoting   the  magnitude  and    direction   of 
the  force  will  be  marked  by  two   capital 
letters,  one  at  each  extremity  ;   while  the 
action-line   will    be    marked  by  the    corre- 
sponding small  letters,  one  being  placed  at 
each  side  of  the  line  designated.     Thus,  in 
Fig.    i,  AB  represents  a  force  in    magni- 
tude and  direction,  while  its  action-line  is  marked  by  the  letters 
ab,  placed  as  shown. 


COMPOSITION    OF   CONCURRENT   FORCES. 


§  2.    Composition  of  Concurrent  Forces. 

13.  Resultant  of  Two  Concurrent  Forces.  —  If  two  concur- 
rent forces  are  represented  in  magnitude  and  direction  by  two 
lines  AB  and  BC,  their 

c, 


(A) 


Fig.  3 


resultant  is  represented 
in  magnitude  and  direc- 
tion by  AC.  (Fig.  2.) 
Proofs  of  this  proposi- 
tion are  given  in  all 
elementary  treatises  on 
mechanics,  and  the  demonstration  will  be  here  omitted.  The 
point  of  application  of  the  resultant  is  the  same  as  that  of  the 
given  force.  Thus  if  O  (Fig.  2)  is  the  given  point  of  applica- 
tion, then  ab,  be,  and  ac,  drawn  parallel  to  AB,  BC,  and  AC 
respectively,  are  the  lines  of  action  of  the  two  given  forces  and 
their  resultant.  The  figure  marked  (A)  is  a  force  diagram,  and 
(B)  is  the  corresponding  space  diagram  (Art.  11). 

14.  Resultant  of  Any  Number  of  Concurrent  Forces.  —  If  any 
number  of  concurrent  forces  are  represented  in  magnitude  and 
direction  by  lines  AB,  BC,  CD,  .  .  .,  their  resultant  is  repre- 
sented in  magnitude  and  direction  by  the  line  AN,  where  N  is 
the  extremity  of  the  line  representing  the  last  of  the  given 
forces. 

This  proposition  follows  immediately  from  the  preceding 
one  ;  for  the  resultant  of  the  forces  represented  by  AB  and 
BC  is  a  force  repre- 
sented by  AC',  the  re- 
sultant of  AC  and  CD 
is  AD,  and  so  on.  By 
continuing  the  process 
we  shall  arrive  at  the 
result  stated.  It  is 

readily  seen  that  the  order  in  which  the  forces  are  taken  does 
not  affect  the  magnitude  or  direction  of  the  resultant  as  thus 


Fig.  3 


6  GRAPHIC    STATICS. 

determined.  The  point  of  application  of  the  resultant  is  the 
same  as  that  of  the  given  forces.  Figure  3  shows  the  force 
diagram  and  space  diagram  for  a  system  of  four  forces  repre- 
sented by  AB,  BC,  CD,  DE,  and  their  resultant  represented 
by  AE,  applied  at  the  point  O. 

It  is  evident  that  every  system  of  concurrent  forces  has  for 
its  resultant  some  single  force  (Art.  9)  ;  though  in  particular 
cases  its  magnitude  may  be  zero. 

15.  Force  Polygon. — The  figure  formed  by  drawing  in  suc- 
cession   lines    representing    in    magnitude   and     direction    any 

number  of  forces  is  called  a  force  polygon  for 
those  forces.  Thus  Fig.  4  is  a  force  polygon 
for  any  four  forces  represented  in  magnitude 
and  direction  by  the  lines  AB,  BC,  CD,  and 
DE,  whatever  their  lines  of  action.  It  may 
happen  that  the  point  E  coincides  with  A,  in 
which  case  the  polygon  is  said  to  be  closed.  It  is  evident  that 
the  order  in  which  the  forces  are  taken  does  not  affect  the 
relative  positions  of  the  initial  and  final  points. 

§  3.    Equilibrium  of  Concurrent  Forces. 

1 6.  Definition. — A  system   of   forces  acting  on  a  body  is 
in  equilibrium  if  the  motion  of  the  body  is  unchanged  by  its 
action. 

17.  Condition   of    Equilibrium.  —  In    order   that    no   motion 
may  result  from  the  action  of  any  system  of  concurrent  forces, 
the  magnitude  of  the  resultant  must  be  zero  ;  and  conversely, 
if    the   magnitude   of    the    resultant    is   zero,    no    motion    can 
result.     But   (Arts.    14   and  15)  the  condition  that  the  result- 
ant  is   zero   is   identical   with   the    condition    that    the    force 
polygon  closes.     Hence,  the  following  proposition  : 

If  any  system  of  concurrent  forces  is  in  equilibrium,  the  force 
polygon  for  the  system  must  close.     And  conversely,  If  the  force 


EQUILIBRIUM    OF   CONCURRENT   FORCES.  7 

polygon  is  closed  for  any  system  of  concurrent  forces ',  the  system 
is  in  equilibrium. 

The    comparison    of   this  with  the   analytical    conditions    of 
equilibrium  is  given  in  Art.  22. 

1 8.  Method  of  Solving  Problems  in  Equilibrium.  —  If   a  sys- 
tem of  concurrent  forces  in  equilibrium  be  partially  unknown, 
we  may  in  certain  cases  determine  the  unknown  elements  by 
applying  the  principles  of  Art.  17. 

The  most  usual  case  is  that  in  which  two  forces  are  unknown 
except  as  to  lines  of  action.     Thus,  suppose  a  system  of  five 
forces    in   equilibrium, 
three        being       fully 
known,  represented  in 
magnitude    and    direc- 
tion   by  AB,  BC,  CD 
(Fig.    5),  and  in    lines 
of  action  by  ab,  be,  cd, 

while  concerning  the  other  two  we  know  only  their  lines  of 
action  de,  ea.  To  determine  these  two  in  magnitude  and 
direction,  it  is  necessary  only  to  complete  the  force  polygon  of 
which  ABCD  is  the  known  part.  The  remaining  sides  must  be 
parallel  respectively  to  de  and  ea.  From  D  draw  a  line  parallel 
to  de,  and  from  A  a  line  parallel  to  ea,  prolonging  them  till  they 
intersect  at  E.  Then  DE  and  EA  represent  the  required  forces 
in  magnitude  and  direction,  and  the  complete  force  polygon  is 
ABCDEA.  It  is  evident  that  ABCDE'A  is  an  equally  legiti- 
mate form  of  the  force  polygon,  and  gives  the  same  result  for 
the  magnitude  and  direction  of  each  of  the  unknown  forces. 
This  problem  occurs  constantly  in  the  construction  of  stress 
diagrams  by  the  method  described  in  Part  II. 

The  student  will  find  little  difficulty  in  treating  other  prob- 
lems in  the  equilibrium  of  concurrent  forces. 

19.  Problems  in  Equilibrium.  —  (i)  A  particle  is  in  equilib- 
rium   under    the    action    of    five    forces,    three   of    which   are 


8  GRAPHIC    STATICS. 

completely  known,  while  of  the  remaining  two,  one  is  known 
in  direction  only,  and  the  other  in  magnitude  only.  To  deter- 
mine the  unknown  forces. 

(2)  Suppose   two   forces    known    in    magnitude    but    not    in 
direction,  the  remaining  forces  being  wholly  known. 

(3)  Suppose   one   force  wholly  unknown,   the   others   being 
known. 

§  4.    Resolution  of  Concurrent  Forces. 

20.  To   Resolve   a   Given   Force  into  Any  Number  of  Com- 
ponents having  the  same  point  of  application,  we  have  only  to 
draw  a  closed  polygon  of  which  one  side  shall  represent  the 
magnitude  and  direction  of  the  given  force  ;  then  the  remaining 
sides  will  represent,  in  magnitude  and  direction,  the  required 
components.     This  problem  is,  in  general,  indeterminate,  unless 
the   components  are  required  to  satisfy  certain  specified  con- 
ditions. 

[NOTE.  —  A  problem  is  said  to  be  indeterminate  if  its  conditions  can  be  satisfied 
in  an  infinite  number  of  ways.  It  is  determinate  if  it  admits  of  only  one  solution. 
Thus,  the  problem,  to  determine  the  values  of  .*•  and  y  which  shall  satisfy  the  equation 
x  +  y  =  10,  is  indeterminate;  while  the  problem,  given  2  x  +  3  =  7,  to  find  the 
value  of  x,  is  determinate.  The  case  in  which  a  finite  number  of  solutions  is  possible 
may  be  called  incompletely  determinate.  Thus,  the  problem,  given  x2  -f  x  —  6  =  o, 
to  find  x,  admits  of  two  solutions,  and  therefore  is  incompletely  determinate.  All 
these  classes  of  problems  may  be  met  with  in  statics.] 

21.  To  Resolve  a  Given  Force  into  Two  Components.  —  This 
problem  is  indeterminate  unless  additional  data  are  given.     For 
if  the  given  force  be  represented  in  magnitude  and  direction  by 
a  line,  any  two  lines  which  with  the  given  line  form  a  triangle 
may  represent  forces  which  are  together  equivalent  to  the  given 
force.     But  an  infinite  number  of  such  triangles  may  be  drawn. 
The  solution  of  the  following  four  cases  of  this  problem  will 
form  exercises  for  the  student.     In  each  case  the  force  diagram 
and  space  diagram  should  be  completely  drawn,  and  the  student 
should  notice  whether  the   problem    is    determinate,   partially 
determinate,  or  indeterminate. 


RESOLUTION   OF   CONCURRENT    FORCES. 


(1)  Let   the   lines  of  action  of  the  required  components  be 
given. 

(2)  Let  the  two  components  be  given  in  magnitude  only.  ^ 

(3)  Let  the  line  of  action  of  one  component  and  the  magni- 
tude of  the  other  be  given. 

(4)  Let  the  magnitude  and  direction  of  one  component  be 
given. 

It  will  be  noticed  that  these  four  cases  correspond  to  four 
cases  of  the  solution  of  a  place  triangle. 

22.  Resolved  Part  of  a  Force.  —  If  a  force  is  conceived  as 
replaced  by  two  components  at  right  angles  to  each  other,  each 
is  called  the  resolved  part*  in  its  direction,  of  the  given  force. 

It  is  readily  seen  that  the  resolved  part  of  a  force  repre- 
sented by  AB  (Fig.  6)  in  the  direction  of  any  line  XX,  is 
represented  in  magnitude  and  direction 
by  A'B',  the  orthographic  projection  of 
AB  upon  XX.  It  follows  that  the 
resolved  part  (in  any  given  direction)  of 
the  resultant  of  any  concurrent  forces 
is  equal  to  the  algebraic  sum  of  the 
resolved  parts  of  its  components  in  that  direction  ;  signs  plus 
and  minus  being  given  to  the  resolved  parts  to  distinguish  the 
two  opposite  directions  which  they 
may  have.  Thus  (Fig.  7)  the  re- 
solved parts  of  the  forces  AB,  BC, 
CD,  in  a  direction  parallel  to  XX, 
are  A '  B' ,  B>  C ,  CD'  ;  and  their 
algebraic  sum  is  A' D' ,  which  is  the 
resolved  part  of  the  resultant  AD. 
If  the  resultant  is  zero,  D'  coincides  with  A' ;  hence, 

(i)  For  the  equilibrium  of  any  concurrent  forces,  the  sum 
of  their  resolved  parts  in  any  direction  must  be  zero. 


Fig.  G 


*  The  term  "resolute"  has  been  proposed  by  J.  B.  Lock  ("  Elementary  Statics  ")  to 
denote  what  is  here  denned  as  the  resolved  part  of  a  force. 


I0  GRAPHIC   STATICS. 

Again,  if  D'  coincides  with  A',  then  either  D  coincides 
with  A,  or  else  AD  is  perpendicular  to  XX ';  hence, 

(2)  If  the  sum  of  the  resolved  parts  of  any  concurrent 
forces  in  a  given  direction  is  zero,  their  resultant  (if  any) 
is  perpendicular  to  that  direction.  And  if  the  sum  of  the 
resolved  parts  is  zero  for  each  of  two  directions,  the  resultant 
is  zero,  and  the  system  is  in  equilibrium. 

Propositions  (i)  and  (2)  state  the  conditions  of  equilibrium 
for  concurrent  forces  usually  deduced  in  treatises  employing 
algebraic  methods. 

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CHAPTER   II.  —  NON-CONCURRENT   FORCES. 

§   i.   Composition  of  Non-concurrent  Forces  Acting  on  the  Same 

Rigid  Body. 

23.  Definition  of  Rigid  Body.  —  A  rigid  body  is  one  whose 
particles  do  not  change  their  positions  relative  to  each  other 
under  any  applied  forces.     No  known  body  is  perfectly  rigid, 
but  for  the  purposes  of  statics,  most  solid  bodies  may  be  con- 
sidered as  such  ;  and  any  body  which  has  assumed  a  form  of 
equilibrium    under   applied   forces,    may,  for   the    purposes    of 
statics,  be  treated  as  a  rigid  body  without  error. 

24.  Change  of  Point  of  Application.  —  The  effect  of  a  force 
upon  a  rigid  body  will  be  the  same,  at  whatever  point  in  its 
line  of  action  it  is  applied,  if  only  the  particle  upon  which  it 
acts  is  rigidly  connected  with  the  body. 

This  proposition  is  fundamental  to  the  development  of  the 
principles  of  statics,  and  is  amply  justified  by  experience.* 
In  applying  the  principle,  we  are  at  liberty  to  assume  a 
point  of  application  outside  the  actual  body,  the  latter  being 
ideally  extended  to  any  desired  limits. 

25.  Resultant   of   Two   Non-Parallel   Forces.  —  If    two  corn- 
planar  forces  are  not  parallel,  their  lines  of  action  must  inter- 
sect, and  the  point  of  intersection  may  be  taken  as  the  point 
of  application  of  each  force.     Hence,  they  may  be  treated  as 


*  This  proposition  may  be  proved  analytically  by  deducing  the  equations  of  motion 
of  a  rigid  body,  and  showing  that  the  effect  of  any  force  on  the  motion  of  the  body 
depends  only  upon  its  magnitude,  direction,  and  line  of  action.  But  such  a  proof  is,  of 
course,  outside  the  scope  of  this  work. 

II 


12  GRAPHIC   STATICS. 

concurrent  forces,  and  their  resultant  may  be  determined  as 
in  Art.  13.  The  following  proposition  may  therefore  be 
stated  : 

If  two  forces  acting  in  the  same  plane  on  a  rigid  body 
are  represented  in  magnitude  and  direction  by  AB  and  BC, 
their  resultant  is  represented  in  magnitude  and  direction  by 

AC,  and  its  line  of  action  passes  through  the  point  of  inter- 
section  of  the  lines  of  action  of   the  given  forces.     Its  point 
of  application  may  be  any  point  of  this  line. 

It  may  happen  that  the  point  of  intersection  of  the  two 
given  lines  of  action  falls  outside  the  limits  available  for  the 
drawing.  In  such  a  case  it  will  be  most  convenient  to  find 
the  resultant  by  the  method  to  be  explained  in  Art.  27.  The 
same  remark  applies  to  the  case  of  two  parallel  forces. 

26.   Resultant   of  Any   Number   of   Non-concurrent   Forces  - 
First  Method.  —  The  method  of  the  preceding  article  may  be 
extended  to  the  determination  of  the  resultant  of  any  number 
of  forces  acting  on  the  same  rigid  body.     Let  AB,  BC,  CD,  DE 

(Fig.  8),  represent  in 
magnitude  and  direction 
four  forces,  and  let  ab, 
be,  cd,  de  represent  their 
lines  of  action.  To  find 
their  resultant,  we  may 
proceed  as  follows : 
The  resultant  of  AB 
and  BC  is  represented 
in  magnitude  and  direction  by  AC,  and  in  line  of  action  by  ac 
drawn  parallel  to  AC  through  the  point  of  intersection  of  ab 
and  be.  Combining  this  resultant  with  CD,  we  get  as  their 
resultant  a  force  represented  in  magnitude  and  direction  by 

AD,  and  in  line  of  action  by  ad  drawn  parallel  to  AD  through 
the  point  of  intersection  of  ac  and  cd.     This  is  evidently  the 
resultant  of  AB,  BC,  and  CD.     In  the  same  way,  this  resultant 


COMPOSITION   OF   NON-CONCURRENT   FORCES.  I3 

combined  with  DE  gives  for  their  resultant  a  force  whose  mag- 
nitude and  direction  are  represented  by  AE,  and  whose  line  of 
action  is  ac,  parallel  to  AE  and  passing  through  the  point  in 
which  ad  intersects  dc.  This  last  force  is  the  resultant  of  the 
four  given  forces. 

The  process  may  evidently  be  extended  to  the  case  of  any 
number  of  forces. 

As  in  the  case  discussed  in  the  preceding  article,  this  method 
will  become  inapplicable  or  inconvenient  in  case  any  of  the 
points  of  intersection  fall  outside  the  limits  available  for  the 
drawing.  For  this  reason  it  is  usually  most  convenient  to 
employ  the  method  described  in  Art.  27. 

The  student  should  bear  in  mind  that  the  length  and  direc- 
tion AE  and  the  line  ae  are  not  the  magnitude,  direction,  and 
line  of  action  of  any  actual  force  applied  to  the  body.  By  the 
resultant  is  meant  an  ideal  force,  which,  if  it  acted,  would 
produce  the  same  effect  upon  the  motion  of  the  body  as  is 
produced  by  the  given  forces.  It  is  a  force  which  may  be 
conceived  to  replace  the  actual  forces,  and  may  be  assumed  to 
be  applied  to  any  particle  in  its  line  of  action,  provided  that 
particle  is  conceived  as  rigidly  connected  with  the  given  body. 
The  line  of  action  may  in  reality  fail  to  meet  the  given  body. 
(See  Art.  24.) 

27.    Resultant  of  Non-concurrent  Forces  —  Second  Method.  - 

This  method  will  be  described  by  reference  to  an  example. 
Referring  to  Fig.  9;  let  AB,  BC,  CD,  DE  represent  in  magni- 
tude and  direction  four  forces  whose  lines  of  action  are  ab,  be, 
cd,  de  ;  and  let  it  be  required  to  find  their  resultant.  Draw  the 
force  polygon  ABCDE,  and  from  any  point  O  in  the  force 
diagram  draw  lines  OA,  OB,  OC,  OD,  OE.  These  lines  may 
represent,  in  magnitude  and  direction,  components  into  which 
the  given  forces  may  be  resolved.  Thus  AB  is  equivalent  to 
forces  represented  by  AO  and  OB  acting  in  any  lines  parallel 
to  AO,  OB,  whose  point  of  intersection  falls  upon  ab\  BC  is 


GRAPHIC    STATICS. 


equivalent  to  forces  represented  by  BO,  OC,  acting  in  any  lines 
parallel  to  BO,  OC,  which  intersect  on  be ;  and  so  for  each  of 
the  given  forces.  The  four  given  forces  may,  therefore,  be 
replaced  by  eight  forces  given  in  magnitude  and  direction  by 


AO,  OB,  BO,  OC,  CO,  OD,  DO,  OE,  with  proper  lines  of 
action.  Now,  it  is  possible  to  make  the  lines  of  action  of  the 
forces  represented  by  OB  and  BO  coincide ;  and  the  same  is 
true  of  each  pair  of  equal  and  opposite  forces,  OC,  CO ;  OD, 
DO.  To  accomplish  this,  let  AO,  OB  act  in  lines  ao,  ob,  inter- 
secting at  any  assumed  point  of  ab.  Prolong  ob  to  intersect 
be,  and  take  the  point  thus  determined  as  the  point  of  inter- 
section of  the  lines  of  action  of  BO,  OC ;  these  lines  are  then 
bo,  oc.  Similarly  prolong  oc  to  intersect  cd,  and  let  the  point 
of  intersection  be  taken  as  the  point  at  which  CD  is  resolved 
into  CO  and  OD ;  the  lines  of  action  of  these  forces  are  then 
co  and  od.  In  like  manner  choose  do,  oe,  intersecting  on  de,  as 
the  lines  of  action  of  DO,  OE.  If  this  is  done,  the  forces  OB, 
BO  will  neutralize  each  other  and  may  be  omitted  from  the 
system  ;  also  the  pairs  OC,  CO,  and  OD,  DO.  Hence,  there 
remain  only  the  two  forces  represented  in  magnitude  and 
direction  by  AO,  OE,  and  in  lines  of  action  by  ao,  oc. 
Their  resultant  is  given  in  magnitude  and  direction  by  AE, 
and  its  line  of  action  is  ae,  drawn  parallel  to  AE  through  the 


COMPOSITION    OF   NON-CONCURRENT   FORCES.  15 

point  of  intersection  of  oa  and  oe ;  and  this  is  also  the  result- 
ant of  the  given  system. 

By  carefully  following  through  this  construction  the  student 
will  be  able  to  reduce  it  to  a  mechanical  method,  which  can 
be  readily  applied  to  any  system. 

x-    ^A^v*1*' C/ 

28.  Funicular  Polygon.  —  The  polygon  whose  sides  are  oa,  ob, 
oc,  od,  oe,  is  called  a.  funicular  polygon*  for  the  given  forces. 

Since  the  point  at  which  the  two  components  of  AB  are 
assumed  to  act  may  be  taken  anywhere  on  the  line  ab,  there 
may  be  any  number  of  funicular  polygons  with  sides  parallel 
to  oa,  ob,  etc.  Again,  if  O  is  taken  at  a  different  point,  there 
may  be  drawn  a  new  funicular  polygon  starting  at  any  point 
of  ab ;  and  by  changing  the  starting  point  any  number  of 
funicular  polygons  may  be  drawn  with  sides  parallel  to  the 
new  directions  of  OA,  OB,  etc.  Moreover,  different  force 
and  funicular  polygons  may  be  obtained  by  changing  the  order 
in  which  the  forces  are  taken. 

It  may  be  proved  geometrically  that  for  every  possible  funic- 
ular polygon  drawn  for  the  same  system  of  forces,  the  last 
vertex,  determined  by  the  above  method,  will  lie  on  the  same 
line  parallel  to  the  closing  side  of  the  force  polygon  (as  ae, 
Fig.  9).  Such  a  proof  is  outside  the  scope  of  this  work. 
The  truth  of  the  proposition  may,  however,  be  shown  from 
the  principles  of  statics.  For  if  it.  were  not  true,  it  would 
be  possible  by  the  above  method  to  find  two  or  more  forces, 
having  different  lines  of  action,  which  are  equivalent  to  each 
other,  because  each  is  equivalent  to  the  given  system.  But 
this  is  impossible. 

29.  Examples.  —  I.   Choose  five  forces,  assigning  the  magni- 
tude,   direction,    and   line    of   action    of   each,    and    find    their 
resultant   by  constructing   the   force   and   funicular    polygons. 

2.  Draw  a  second  funicular  polygon,  using  the  same  point 
O  in  the  force  diagram. 

*Also  called  equilibrium  polygon. 


T6  GRAPHIC    STATICS. 

3.  Draw  a  third  funicular  polygon,  choosing  a  new  point  O. 

4.  Solve  the  same  problem,  taking  the  forces  in  a  different 
order. 

30.  Definitions.  —  The  point  O  (Fig.  9)  is  called  the  pole  of 
the  force   polygon.     The   lines    drawn    from    the    pole   to    the 

•vertices  of  the  force  polygon  may  be  called  rays.  The  sides 
of  the  funicular  polygon  are  sometimes  called  strings. 

Each  ray  in  the  force  diagram  is  parallel  to  a  corresponding 
string  in  the  space  diagram.  As  a  mechanical  rule,  it  should 
be  remembered  that  the  two  rays  draivn  to  the  extremities  of  the 
line  representing  any  force  are  respectively  parallel  to  the  two 
strings  which  intersect  on  the  line  of  action  of  that  force. 

The  rays  terminating  at  the  extremities  of  any  side  of  the 
force  polygon  represent  in  magnitude  and  direction  two  com- 
ponents that  may  replace  the  force  represented  by  that  side ; 
while  the  corresponding  strings  represent  the  lines  of  action  of 
these  components.  Thus  (Fig.  9)  the  force  represented  by  BC 
may  be  replaced  by  two  forces  represented  by  BO,  OC,  acting 
in  the  lines  bo,  oc. 

The  pole  distance  of  any  force  is  the  perpendicular  distance 
from  the  pole  to  the  line  representing  that  force  in  the  force 
diagram.  It  may  evidently  be  considered  as  representing  the 
resolved  part,  perpendicular  to  the  given  force,  of  either  of  the 
components  represented  by  the  corresponding  rays.  Thus,  OM 
(Fig.  9)  is  the  pole  distance  of  AB  ;  and  OM  represents  the 
resolved  part,  perpendicular  to  AB,  of  either  OA  or  OB.  The 
student  should  notice  particularly  that  the  pole  distance  repre- 
sents a  force  magnitude  and  not  a  length. 

31.  Forces  not  Possessing  a  Single  Resultant.  —  It  may  hap- 
pen that  the  first  and  last  sides  of  the  funicular  polygon  are 
parallel,  so  that  the  above  construction  fails  to  give  the  line  of 
action  of  the  resultant.     This  will  be  the  case  if  the  pole  is 
chosen  on  the  line  AE  (Fig.  9),  because  the  first  and  last  sides 
of  the  funicular   polygon   are  respectively  parallel  to  OA  and 


*      V  *r<r°       (^VNj*v      \K&     'V'v 

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VC^>       <Nv^V  ^^    ^ 

I) 

COMPOSITION    OF   NON-CONCURRENT   FORCES.  j^ 

f^^  &fc    ^Waj^-vo    "^WX     'VxA 

OE.  The  difficulty  will,  in  this  case,  be  avoided  by  taking  the 
pole  at  some  point  not  on  the  line  AE.  But  in  one  particular 
case  OA  and  OE  will  be  parallel,  wherever  the  pole  be  taken. 
By  inspection  of  the  force  diagram  it  is  seen  that  this  can 
occur  only  when  the. point  E  coincides  with  A.  In  this  case 
AO  and  OE  represent  equal  and  opposite  forces  ;  and  unless 
their  lines  of  action  coincide,  they  cannot  be  combined  into  a 
simpler  system.  If  their  lines  of  action  are  coincident,  the 
forces  neutralize  each  other  and  the  resultant  is  zero  ;  if  not, 
the  system  reduces  to  a  couple  (Art.  7).  If  the  lines  of  action 
of  AO  and  OE  coincide,  they  may  still  be  regarded  as  forming 
a  couple,  its  arm  being  zero.  Therefore, 

If  the  force  polygon  for  any  system  of  forces  closes,  the  result- 
ant is  a  couple. 

Now,  it  is  evident  that  by  shifting  the  starting  point  for  the 
funicular  polygon,  the  lines  of  action  of  AO  and  OE  will  be 
shifted ;  and  by  taking  a  new  pole,  their  magnitude,  or  direc- 
tion, or  both,  may  be  changed.  Hence,  there  may  be  found 
any  number  of  couples,  each  equivalent  to  the  same  system 
of  forces,  and  therefore  equivalent  to  each  other.  The  relation 
between  equivalent  couples  is  discussed  in  Art.  52. 

Example. — Assume  five  forces  whose  force  polygon  closes, 
the  lines  of  action  being  taken  at  random.  Draw  two  funicular 
polygons,  using  the  same  pole,  and  a  third,  using  a  different 
pole ;  thus  reducing  the  given  system  to  an  equivalent  couple 
in  three  different  ways. 

32.  Resultant  Force  and  Resultant  Couple.  —  From  the  above 
discussion  (Arts.  27  and  31),  it  is  evident  that  any  system  of 
complanar  forces,  acting  on  the  same  rigid  body,  is  equivalent 
either  to  a  single  force  or  to  a  couple.     In  other  words,  every 
system  of  complanar  forces  possesses  either  a  resultant  force  or  a 
resultant  couple.     (See  Art.  9.) 

33.  Comparison  of  Methods.  —  Of  the  methods  given  in  Arts. 
26  and  27,  for  finding  the  resultant  of  a  system  of  non-concur- 

-  vc~o—       vV   V^'-'^J^-tL--      \X]     ^*§j^j\3is*\s*3\/*k 
v\  » 

VVvvX/v-^jJ^ 


jg  GRAPHIC    STATICS. 

rent  forces,  the  first  is  a  special  case  of  the  second.  For  if  the 
pole  in  Fig.  9  be  chosen  at  the  point  A,  and  the  first  string  be 
made  to  pass  through  the  point  of  intersection  of  ab  and  be,  the 
construction  becomes  identical  with  that  of  Fig.  8.  If  the  first 
method  is  employed,  we  are  liable  to  meet  the  difficulty  men- 
tioned in  Arts.  25  and  26,  that  some  of  the  required  points  of 
intersection  do  not  fall  within  convenient  limits. 

In  the  second  method,  since  the  pole  may  be  chosen  at  pleas- 
ure, the  rays  may  generally  be  caused  to  make  convenient  angles 
with  the  given  forces.  It  will  rarely  be  necessary  that  the  rays 
intersect  the  corresponding  force  lines  at  acute  angles,  and  will 
never  be  necessary  that  a  ray  shall  be  parallel  to  the  force  of 
which  it  represents  a  component.  Hence  the  pole  may  gener- 
ally be  so  chosen  that  all  intersections  shall  fall  within  conven- 
ient limits.  The  difficulty  above  mentioned  may  therefore 
usually  be  avoided  by  using  the  method  last  described.  This 
method  is  especially  convenient  in  the  case  of  parallel  forces. 

Example.  —  Choose  six  parallel  forces  and  find  their  resultant 
by  constructing  the  force  and  funicular  polygons. 

34.  Closing  of  the  Funicular  Polygon.  —  Let  the  force  poly- 
gon be  drawn  for  any  given  forces,  and  let  A  and  E  be  the 
initial  and  final  points.     Suppose  the  funicular  polygon  drawn, 
corresponding  to  any  pole  O.     The  given  system  is  equivalent 
to  two  forces  represented  in  magnitude  and  direction  by  AO, 
OE,  their  lines  of  action  being  the  first  and  last   sides  of  the 
funicular  polygon.     In  general,  these  lines  of  action  will  not  be 
parallel;  but,  as  explained  in  Art.  31,  it  may  happen  that  they 
are  parallel.     And  if  parallel,  it  may  happen  that  they  coincide. 
In  this  case,  the  funicular  polygon  is  said  to  be  closed. 

§  2.  Equilibrium  of  Non-concurrent  Forces. 

35.  Conditions  of  Equilibrium.  —  From  the  preceding  articles 
it  follows  that,  in  order  that  a  system  of  complanar  forces  may 
be  in  equilibrium,  the  following  conditions  must  be  satisfied : 


EQUILIBRIUM    OF   NON-CONCURRENT   FORCES.  ICj 

(a)  The  force  polygon  must  close  ;  otherwise  the  construction  of 
Art.  27  will  always  lead  to  a  resultant  force  of  finite  magnitude. 

(6)  Every  funicular  polygon  must  close ;  otherwise,  by  the 
method  of  Art.  27,  the  system  can  be  reduced  to  two  forces  with 
different  lines  of  action  ;  the  resultant  cannot,  therefore,  be  zero. 

Conversely,  if  tJie  force  polygon  is  closed  and  one  funicular 
polygon  also  closes,  tJie  system  is  in  equilibrium.  For,  the  clos- 
ing of  the  force  polygon  shows  that  the  system  may  be  reduced 
to  two  equal  and  opposite  forces  (Art.  31)  ;  and  the  closing  of 
the  funicular  polygon  shows  that  these  have  the  same  line  of 
action,  and  therefore  balance  each  other. 

It  follows  that  if  the  force  polygon  and  one  funicular  polygon 
are  closed,  all  funicular  polygons  will  close. 

36.  Auxiliary  Conditions  of  Equilibrium.  —  If  any  system  of 
forces  in  equilibrium  be  divided  into  two  groups,  the  resultants 
of  the  two  groups  are  equal  and  opposite  and  have  the  same 
line  of  action. 

Particular  case.  —  If  three  forces  are  in  equilibrium,  their 
lines  of  action  must  meet  in  a  point,  or  be  parallel.  For,  if 
the  lines  of  action  of  two  of  the  forces  intersect,  their  result- 
ant must  act  in  a  line  passing  through  the  point  of  inter- 
section. But  this  resultant  must  be  equal  and  opposite  to 
the  third  force  and  have  the  same  line  of  action. 

37.  General  Method  of  Solving  Problems  in  Equilibrium.  - 

The  principles  of  Art.  35  furnish  a  general  method  of  solv- 
ing, graphically,  problems  relating  to  the  equilibrium  of  corn- 
planar  forces  acting  on  a  rigid  body.  The  problems  to  be 
solved  will  always  be  of  the  following  kind  : 

A  body  is  in  equilibrium  under  the  action  of  forces,  some 
of  which  are  completely  known,  and  others  wholly  or  partly 
unknown.  It  is  required  to  determine  the  unknown  elements. 
The  required  elements  may  be  either  the  magnitudes  or  the 
lines  of  action  of  the  forces. 


2O 


GRAPHIC    STATICS. 


The  general  method  of  procedure  is  as  follows  :  First,  draw 
the  force  polygon  and  funicular  polygon  so  far  as  possible 
from  the  given  data ;  then  complete  them,  subject  to  the 
condition  that  both  polygons  must  close. 

This  general  method  will  be  illustrated  by  the  solution  of 
several  problems  of  frequent  occurrence.  We  may  here  meet 
with  both  determinate  and  indeterminate  problems  (Art.  20). 
In  order  that  a  problem  may  be  determinate,  the  given  data, 
together  with  the  condition  that  the  force  polygon  and  funic- 
ular polygon  are  to  close,  must  be  just  sufficient  to  fully 
determine  these  figures.  There  are  many  possible  cases  fur- 
nishing determinate  problems.  Some  of  these  cannot  readily 
be  solved  graphically.  In  the  following  articles  are  treated 
three  important  cases  to  which  the  general  method  above 
outlined  is  well  adapted. 

38.  Problems  in  Equilibrium.  I.  —  A  rigid  body  is  in  equi- 
librium under  the  action  of  a  system  of  parallel  forces,  all 
known  except  two,  these  being  unknown  in  magnitude  and 
direction,  but  having  known  lines  of  action.  It  is  required 
to  fully  determine  the  unknown  forces. 

Let  the  known  lines  of  action  of  five  forces  in  equilibrium 

be  ab,  be,  cd,  de,  ca  (Fig.  10),  and  let 
AB,  BC,  CD  represent  the  known 
forces  in  magnitude  and,  direction, 
while  DE  and  EA  are  at  first  un- 
known. The  force  polygon,  so  far 
as  it  can  be  drawn  from  the  given 
data,  is  the  straight  line  ABCD. 
Choose  a  pole  O  and  draw  rays  OA, 
OB,  OC,  OD.  Parallel  to  these  draw 
in  the  space  diagram  the  strings  oa, 
ob,  oc,  od,  four  successive  sides  of  the 
funicular  polygon.  This  much  can 
be  drawn  from  the  data  given.  We  must  now  complete  the 


aft 


be     a 


d     d 


Fig.  1O 


EQUILIBRIUM   OF   NON-CONCURRENT   FORCES. 


21 


two  polygons  and  cause  both  to  close.  In  the  funicular  poly- 
gon, but  one  side  remains  to  be  drawn ;  and  in  the  force 
polygon  but  one  vertex  remains  unknown.  If  the  forces  are 
taken  in  the  order  AB,  BC,  CD,  DE,  EA,  the  unknown  vertex 
in  the  force  polygon  is  the  one  to  be  marked  E ;  the  unknown 
side  of  the  funicular  polygon  is  oe,  and  is  to  be  parallel  to 
the  line  OE.  But  the  string  oe  must  pass  through  the  inter- 
section of  od  with  de,  and  also  through  the  intersection  of 
oa  with  ac.  Hence,  since  these  intersections  are  both  known, 
oe  can  be  drawn  at  once  as  shown  in  the  figure ;  and  then 
OE  can  be  drawn  parallel  to  oe.  This  fixes  the  point  E ; 
and  DE  and  EA  represent,  in  magnitude  and  direction,  the 
required  forces  whose  lines  of  action  are  de,  ea. 

39.  Problems  in  Equilibrium.  II.  —  A  rigid  body  is  in 
equilibrium  under  the  action  of  a  system  of  non-parallel  forces, 
all  known,  except  two ;  of  these,  the  line  of  action  of  one 
and  the  point  of  application  of  the  other  are  given.  It  is 
required  to  completely  determine  the  unknown  forces. 

Let  the  system  consist  of  five  forces  to  be  represented  in 
the  usual  way  ;  in  magnitude  and  direction  by  lines  marked 
AB,  BC,  CD,  DE,  EA  ;  and  in  lines  of  action  by  ab,  be,  cd, 
de,  ea.  Of  these  lines  let  be,  cd,  de,  ea  (Fig.  n),  be  given,  and 
let  L  be  a  given  point  of  ab.  Also  let  BC,  CD,  and  DE  be 
known,  while  EA,  AB  are  unknown.  First,  draw  the  force 
polygon  as  far  as  possible,  giving  B,  C,  D,  E,  four  consecutive 
vertices  of  the  force 
polygon.  The  side  EA 
must  be  parallel  to  ca, 
but  its  length  is  un- 
known, hence  the  ver- 
tex A  of  the  force 
polygon  cannot  be 
fixed.  Next,  draw  the 
funicular  polygon  so 


Fig.  11 


22  GRAPHIC    STATICS. 

far  as  possible  from  the  given  data.  Choose  the  pole  O,  and 
draw  the  rays  OB,  OC,  OD,  OE.  The  remaining  ray,  OA, 
cannot  yet  be  drawn.  Now,  in  the  funicular  polygon,  the 
sides  ob,  oc  must  intersect  on  be;  oc  and  od  must  intersect 
on  cd;  od  and  oe  must  intersect  on  de ;  oe  and  oa  must  inter- 
sect on  ca  ;  and  oa  and  ob  must  intersect  on  ab.  Since  L 
is  the  only  known  point  of  the  line  ab,  let  this  point  be 
taken  as  the  point  of  intersection  of  oa  and  ob.  Now  draw 
ob  through  L  parallel  to  OB,  and  successively  oc,  od,  oe,  par- 
allel to  OC,  OD,  OE.  We  may  now  draw  oa,  joining  L  with 
the  point  in  which  oe  intersects  ea.  This  completes  the  funic- 
ular polygon.  The  force  polygon  can  now  be  completed  as 
follows  :  From  O  draw  a  line  parallel  to  oa  and  from  E  a  line 
in  the  known  direction  of  EA  ;  their  intersection  determines 
A.  This  determines  both  EA  and  AB,  and  the  force  polygon 
is  completely  known.  The  line  of  action  ab  may  now  be 
drawn  through  L  parallel  to  AB,  and  the  forces  are  completely 
determined. 

40.  Problems  in  Equilibrium.  III. — A  rigid  body  is  in 
equilibrium  under  the  action  of  any  number  of  forces,  of 
which  three  are  known  only  in  line  of  action ;  the  remaining 
forces  being  completely  known.  It  is  required  to  determine 
the  unknown  forces. 

Let  the  given  forces  be  six  in  number,  their  lines  of  action 
being  represented  by  ab,  be,  cd,  de,  ef,  fa  (Fig.  12),  and  let  AB, 
BC,  and  CD  be  known,  while  the  remaining  three  forces  are  un- 
known in  magnitude  and  direction.  The  known  data  make  it 
possible  to  draw  at  once  three  sides  of  the  force  polygon, 
namely,  AB,  BC,  CD;  and  the  four  rays  OA,  OB,  OC,  OD, 
from  any  assumed  pole  O.  Also,  four  sides  of  the  funicular 
polygon,  oa,  ob,  oc,  od,  may  be  drawn  at  once.  But  oc  and  of 
are  unknown  ;  also  DE,  EF,  and  FA  in  the  force  polygon. 

Since  any  side  of  the  funicular  polygon  can  be  drawn  through 
any  chosen  point,  let  the  polygon  be  started  by  drawing  oa 


EQUILIBRIUM    OF   NON-CONCURRENT    FORCES.  23 

through  the  intersection  of  efaud  fa.  We  may  then  draw  suc- 
cessively ob,  oc,  od.  Now,  of  is  unknown  in  direction  ;  but  it 
is  to  be  drawn  through  the  intersection  of  oa  and  fa,  hence, 
whatever  its  direction,  it  intersects  ef  in  the  same  point  (since 
oa  was  drawn  through  the  intersection  of  ef  and  fa).  Hence, 
the  two  vertices  of  the  funicular  polygon  falling  on  ef  and  fa 
coincide  at  the  intersection  of  these  two  lines.  We  may  there- 


fore  draw  oe  through  this  point  and  also  through  the  point 
already  found  by  the  intersection  of  od  and  de.  Now  draw  a 
line  from  O  parallel  to  oe,  and  from  D  a  line  parallel  to  de ; 
their  intersection  is  E.  Draw  from  E  a  line  parallel  to  ef,  and 
from  A  a  line  parallel  to  fa ;  their  intersection  determines  F. 
The  force  polygon  is  now  completely  drawn,  and  DE,  EF,  FA 
represent,  in  magnitude  and  direction,  the  required  forces.  The 
remaining  ray  OF  and  the  corresponding  string  of  may  now  be 
drawn,  but  are  not  needed. 

The  construction  might  have  been  made  equally  well  by 
choosing  the  intersection  of  de  and  ef  as  the  starting  point, 
since  two  vertices  of  the  funicular  polygon  may  be  made  to 
coincide  at  that  point.  Or,  the  point  of  intersection  of  de  and 
of  might  be  chosen ;  •  but  in  that  case,  the  order  of  the  forces 
should  be  so  changed  as  to  make  de  and  af  consecutive.  If 
this  were  done  the  figures  should  be  relettered  to  agree  with 


GRAPHIC   STATICS. 


the  order  in  which  the  forces  were  taken.  It  may  be  noticed 
that  the  direction  of  the  string  first  drawn  may  be  chosen  arbi- 
trarily and  the  pole  so  taken  as  to  correspond  to  the  direction 
chosen.  This  is  important  in  the  treatment  of  the  following 
special  case. 

Case  of  inaccessible  points  of  intersection.  —  It  may  happen 
that  the  lines  of  action  dc,  ef,  and  fa  have  no  point  of  intersec- 
tion within  convenient  limits.  When  this  is  the  case,  the 
method  just  given  may  still  be  applied,  but  involves  the  geo- 
metrical problem  of  drawing  a  line  through  an  inaccessible  point. 
For  example,  if  ef  and  fa  intersect  beyond  the  limits  of  the 
drawing,  as  shown  in  Fig.  13,  we  may  proceed  as  follows: 
Choose  some  point  of  ab  and  from  it  draw  a  line  through  the 
point  of  intersection  of  ef  and  fa.  (This  can  be  done  by  a 
jnethod  to  be  explained  presently.)  Let  this  line  be  oa.  From 
the  known  point  A  of  the  force  polygon  draw  a  line  parallel  to 


Fig.  13 


oa  and  choose  a  pole  upon  it.  Draw  the  rays  OB,  OC,  OD ; 
then  the  corresponding  strings  in  the  order  ob,  oc,  od.  From 
the  point  in  which  od  intersects  de  draw  a  line  to  the  inaccessible 
point  of  intersection  of  ef  and  fa  ;  this  will  be  oe.  The  force 
polygon  can  now  be  completed  just  as  in  the  preceding  case. 

A  line  may  be  drawn  through  the  inaccessible  point  of  inter- 
•  section  of  two  given  lines  by  the  following  method  :  Let  AC, 


EQUILIBRIUM    OF   NON-CONCURRENT   FORCES. 


(Fig.  14),  be  the  given  lines,  and  let  it  be  required  to  draw 
a  line  through  their  point  of  intersection  from  some  point  P. 
Draw  PA  intersecting  AC  and 
BD,  and  from  C,  any  point  of 
AC,  draw  a  line  CQ  parallel  to 
PA.  From  E,  a  point  of  BD, 
draw  EA,  EP  ;  also  draw  CF 
parallel  to  AE,  and  FQ  parallel 
to  EP.  Then  PQ  is  the  line 
required.  For,  by  the  similar 
triangles,  it  is  easily  shown  that 

'—- ;  which  proves  that  AC,  BG,  and  PQ  meet  in  a  point. 
GQ 

Exception.  —  If  the  lines  of  action  of  the  three  unknown 
forces  meet  in  a  point  (or  are  parallel),  the  problem  is  impos- 
sible of  solution,  unless  it  happens  that  the  resultant  of  the 
known  forces  acts  in  a  line  through  the  point  of  intersection 
of  those  three  lines  of  action  (or  parallel  to  them)  ;  in  which 
case  the  problem  is  indeterminate.  These  cases  will  not  be 
discussed  here. 

Remark.  —  In  the  problems  treated  in  this  and  the  two  pre- 
ceding articles,  it  will  be  noticed  that  the  forces  should  be  taken 
in  such  an  order  that  those  which  are  completely  known  are 
consecutive.  Otherwise  the  number  of  unknown  lines  in  the 
force  and  funicular  polygons  will  be  increased.  [For  an- 
other method  of  solving  this  problem,  see  Levy's  "  Statique 
Graphique."] 

41.  Examples.  —  I.  A  rigid  beam  AB  rests  horizontally 
upon  supports  at  A  and  B,  and  sustains  loads  as  follows : 
Its  own  weight  of  100  Ibs.  acting  at  its  middle  point  ;  a  load 
of  50  Ibs.  at  C ;  a  load  of  60  Ibs.  at  D ;  and  a  load  of  80  Ibs. 
at  E.  The  successive  distances  between  A,  C,  D,  E,  and  B 
are  4  ft.,  6  ft.,  7  ft.,  and  10  ft.  Required  the  upward  pressures 
on  the  beam  at  the  supports. 


e 


03 


26  GRAPHIC    STATICS. 

2.  A  rigid  beam  AB  is  hinged  at  A,  and  rests  horizontally 
with  the  end  B  upon  a  smooth  horizontal  surface.     The  beam 
sustains  loads  as  in  Example  i,  and  an  additional  force  of  40- 
Ibs.  is  applied  at  the  middle  point  at  an  angle  of  45°  with  the 
bar  in  an  upward  direction.     Required  the  pressures  upon  the 
beam  at  A  and  B.     [The  pressure  at  A  may  have  any  direc- 
tion, while  the  pressure  at  B  must  be  vertically  upward,  i.e., 
at  right  angles  to  the   supporting  surface.     Hence,  this  is   a 
particular  case  of  Problem  II.] 

3.  Let  the  end  B  of  the  bar  rest  against  a  smooth  surface 
making   an   angle  of  30°  with    the    horizontal ;  the   remaining 
data  being  as  in  Example  2. 

4.  A  rigid  bar  2  ft.  long  weighs  10  Ibs.,  its  center  of  gravity 
being  8  inches  from  one  end.     The  bar  rests  inside  a  smooth 
hemispherical  bowl    of    15  inches  radius.     What  weight  must 
be  applied  at  the  middle  point  in  order  that  the  bar  may  rest 
when    making   an    angle   of    15°   with    the   horizontal?     Also, 
what  are  the  reactions  at  the  ends  ? 

5.  A  uniform  bar  20  inches    long,   weighing   10  Ibs.,    rests 
with   one  end   against  a  smooth   vertical  wall   and  the  other 
end  overhanging  a  smooth  peg    10  inches  from  the  wall.     A 
weight  P  is  suspended  from  the  end  so  that    the    bar   is    in 
equilibrium  when  making  an  angle  of  30°  with  the  horizontal. 
Find  P,  and  the  pressures  exerted  on  the   bar   by   the   wall 
and  peg. 

42.  Special  Methods.  —  Certain  problems  can  be  treated 
more  simply  by  other  methods  than  by  the  general  method 
of  constructing  the  force  and  funicular  polygons.  This  is 
sometimes  true  of  the  following  : 

Problem.  —  A  rigid  body  is  held  in  equilibrium  by  four 
forces  acting  in  known  lines,  only  one  being  known  in  magni- 
tude and  direction.  It  is  required  to  completely  determine 
the  three  remaining  forces.  (See  Fig.  15.) 

Let  the  four  forces  have  lines  of  action  ab,  be,  cd,  da,  and  let 


EQUILIBRIUM    OF    NON-CONCURRENT   FORCES.  2J 

AB  be  drawn  representing  the  magnitude  and  direction  of  the 
known  force.  Now  the  resultant  of  the  forces  whose  lines 
of  action  are  da  and  ab  must  act  in  a  line  passing  through 
Mt  the  point  of  intersection  of  these  lines  ;  and  the  resultant 
of  the  other  two  forces  must  act  in  a  line  passing  through 
N,  the  point  of  intersection  of  be  and  cd.  But  the  two 


resultants  must  be  equal  and  opposite  and  have  the  same 
line  of  action,  else  there  could  not  be  equilibrium.  (Art.  36.) 
Hence,  each  must  act  in  the  line  MN..  Draw  BD  parallel 
to  MNt  and  AD  parallel  to  ad;  the  point  D  being  determined 
by  their  intersection.  Then  DA  represents  in  magnitude  and 
direction  the  force  acting  in  da,  and  DB  the  resultant  of 
DA  and  AB.  But  BD  must  represent  the  resultant  of  the 
two  remaining  forces  ;  hence  these  two  forces  are  represented 
by  BC  and  CD  drawn  from  B  and  D  parallel  respectively  to 
be  and  cd. 

This  problem  is  a  special  case  of  that  treated  in  Art.  40. 
But  the  construction  here  given  will  in  many  cases  be  the 
simpler  one. 

Example.  —  A  rigid  body  has  the  form  of  a  square  ABCD, 
the  side  AB  being  horizontal,  and  BC  vertical.  The  weight 
of  the  body  is  100  Ibs.,  its  center  of  gravity  being  at  the  inter- 
section of  the  diagonals.  It  is  held  in  equilibrium  by  three 
forces  as  follows  :  P1  acting  at  C  in  the  line  AC  '  ;  P>2  acting  at 
D  in  the  line  AD  ;  and  P3  applied  at  B  and  acting  In  the  line 
joining  B  with  the  middle  point  of  AD.  RequVJtto  com- 
pletely determine  Pl}  P2,  and  P3. 


A 


V 


23  GRAPHIC    STATICS. 

§  3.    Resolution  into  Non-concurrent  Systems. 

43.  To  Replace   a   Force  by  Two  Non-concurrent  Forces.  - 

This  may  be  done  in  an  infinite  number  of  ways.  The  lines 
of  action  of  the  two  components  must  intersect  at  a  point  on 
the  line  of  action  of  the  given  force,  and  they  must  further 
satisfy  the  same  conditions  as  concurrent  forces.  (Art.  21.) 

44.  To  Replace  a  Force  by  More  Than  Two  Non-concurrent 
Forces.  —  This  may  be  done  by  first  resolving  the  given  force 
into  two  forces   by  the  preceding  article,  and   then   resolving 
one  or  both  of  its  components  in  the  same  way.     This  problem 
and  that   of  Art.  43   are   indeterminate.     (See   note,  Art.   20.) 
To    make    such   a   problem    determinate,   something   must    be 
specified  regarding  the  magnitudes  and  lines  of  action  of  the 
required  components.     We  shall  consider  some  of  the  particular 
cases  which  are  of  frequent  use  in  the  treatment  of  practical 
problems. 

45.  Resolution  of  a  Force  into  Two  Parallel  Components.  — 

Let  it  be  required  to  resolve  a  force  into  two  components 
having  given  lines  of  action  parallel  to  its  own. 

If  the  given  force  be  reversed  in  direction,  it  will  form  with 
the  required  components  a  system  in  equilibrium.  The  com- 

ponents may  then  be  determined  by  the  method  of  Art.  38. 
v  (5 

Example.  —  Let    the    student   solve   two  particular  cases   of 

• 

this  problem,  taking  the  line  of  action  of  the  given  force  (i) 
between  those  of  the  components,  (2)  outside  those  of  the 
components. 

46.  Resolution  of  a  Force  into  Three  Components.  —  Problem. 
-To  resolve  a  force  into  three  components  having  known  lines 

of  action. 

If  the  given  force  be  reversed  in  direction,  it  will  form,  with 
the  required  forces,  a  system  in  equilibrium.  Hence  these 
forces  may  be  determined  by  either  of  the  two  methods  given 


RESOLUTION    INTO   NON-CONCURRENT    SYSTEMS. 


29 


in    Arts.    40   and    42.      Or,    the    following    reasoning    may    be 
employed,  leading  to  the  same  construction  as  that  of  Art.  42. 

Let  AD  (Fig.  16)  represent  the  given  force  in  magnitude 
and  direction,  and  ad  its  line  of  action  ;  the  lines  of  action  of 
the  components  being  given  as  ab,  be,  cd.  Since  the  given 
force  may  be  assumed  to  act  at  any  point  in  the  line  ad,  let  its 
point  of  application  be  taken  at  M,  the  point  of  intersection  of 


Fig.  16 


ad  and  cd.  Resolve  it  into  two  components  acting  in  the  lines 
cd  and  MN,  N  being  the  point  of  intersection  of  ab  and  be. 
These  components  are  represented  in  magnitude  and  direction 
by  AC,  CD.  Let  AC,  acting  in  the  line  MN  (also  marked  ac), 
be  resolved  into  components  having  lines  of  action  ab,  be. 
These  components  are  given  in  magnitude  and  direction  by 
AB,  BC,  drawn  parallel  respectively  to  ab,  be.  Hence,  the 
given  force,  acting  in  ad,  is  equivalent  to  the  three  forces 
represented  in  magnitude  and  direction  by  AB,  BC,  CD,  acting 
in  the  lines  ab,  be,  cd. 

If  the  line  of  action  of  the  given  force  does  not  intersect  any 
one  of  the  given  lines  ab,  be,  cd,  within  the  limits  of  the  draw- 
ing, it  may  be  replaced  by  two  components  ;  then  each  may  be 
resolved  in  accordance  with  the  above  method,  and  the  results 
combined.  If  the  three  lines  ab,  be,  cd,  are  all  parallel  to  ad, 
or  if  the  four  lines  intersect  in  a  point,  the  problem  is  inde- 
terminate. This  is  evident  from  the  preceding  articles,  since, 
by  methods  already  given,  a  part  of  AD  can  be  replaced  by 
two  components  acting  in  any  two  of  the  given  lines,  as  ab, 
be ;  and  the  remaining  part  by  two  components  acting  in  ab,  cd; 


30  GRAPHIC    STATICS. 

or  in   be,   cd.     This   construction   evidently   admits    of   infinite 
variation. 

For  another  method  of  solving  the  above  problem,  see  Clarke's 
"  Graphic  Statics,"  p.  16. 

§  4.    Moments  of  Forces  and  of  Couples. 

47.  Moment  of  a  Force.  —  Definition.  —  The  moment  of  a 
force  with  respect  to  a  point  is  the  product  of  the  magnitude  of 
the  force  into  the  perpendicular  distance  of  its  line  of  action 
from  the  given  point.  The  moment  of  a  force  with  respect  to 
an  axis  perpendicular  to  the  force  is  the  product  of  the  magni- 
tude of  the  force  by  the  perpendicular  distance  from  the  axis 
to  the  line  of  action  of  the  force. 

If  the  moment  is  taken  with  respect  to  a  point,  that  point  is 
called  the  origin  of  moments.  The  perpendicular  distance  from 
the  origin,  or  axis,  to  the  line  of  action  of  the  force  is  called 
the  arm. 

Rotation  tendency  of  a  force.  —  The  moment  of  a  force 
measures  its  tendency  to  produce  rotation  about  the  origin,  or 
axis.  Thus,  if  a  rigid  body  is  fixed  at  a  point,  but  free  to  turn 
about  that  point  in  a  given  plane,  any  force  acting  upon  it  in 
that  plane  will  tend  to  cause  it  to  rotate  about  the  fixed  point. 
The  amount  of  this  tendency  will  be  proportional  both  to  the 
magnitude  of  the  force  and  to  the  distance  of  its  line  of  action 
from  the  given  point  ;  that  is,  to  the  moment  of  the  force  with 
respect  to  the  point,  as  above  defined. 

Rotation  in  any  plane  may  have  either  of  two  opposite 
directions,  which  may  be  distinguished  from  each  other  by  signs 
plus  and  minus.  Rotation  with  the  hands  of  a  watch  supposed 
placed  face  upward  in  the  plane  of  the  paper  will  be  called 
negative,  and  the  opposite  kind  positive.  It  would  be  equally 
legitimate  to  adopt  the  opposite  convention,  but  the  method 
here  adopted  agrees  with  the  usage  of  the  majority  of  writers. 

The  sign  of  the  moment  of  a  force  is  regarded  as  the  same 
as  that  of  the  rotation  it  tends  to  produce  about  the  origin. 


MOMENTS   OF   FORCES   AND   OF   COUPLES.  3! 

Moment  represented  by  the  area  of  a  triangle.  —  If  a  triangle 
be  constructed  having  for  its  vertex  the  origin  of  moments,  and 
for  its  base  a  length  in  the  line  of  action  of  a  force,  numerically 
equal  to  its  magnitude,  then  the  moment  of  the  force  is  numeri- 
cally equal  to  double  the  area  of  the  triangle.  This  follows  at 
once  from  the  definition  of  moment. 


48.  Moment  of  Resultant  of  Two  Non-parallel  Forces.  —  Propo- 
sition. —  The  moment  of  the  resultant  of  two  non-parallel  forces 
with  reference  to  a  point  in  their  plane  is  equal  to  the  algebraic 
sum  of  their  separate  moments  with  reference  to  the  same  point. 

In  Fig.  17,  let  AB,  BC,  and  AC  represent  in  magnitude  and 
direction  two  forces  and  their  resultant  ;  and  let  ab,  be,  ac  be 
their  lines  of  action,  intersecting  in  a 
point  N.  Choose  any  point  M  as  the 
origin  of  moments.  Lay  off  NP  =  AB\ 
NQ  =  BC\  and  NR  =  AC.  Then  the 
moments  of  the  three  forces  are  re- 
spectively equal  to  double  the  areas  of 
the  triangles  MNP,  MNQ,  MNR. 
These  three  triangles  have  a  common 
side  MN,  which  may  be  considered  the 
base  ;  hence  their  areas  are  propor- 
tional to  their  altitudes  measured  from  that  side.  These  alti- 
tudes are  PP1,  QQ',  RR1  ',  perpendicular  to  MN.  Now,  if  PP" 
is  parallel  to  MN,  P"R'  is  equal  to  PPf  ;  also  RP"  equals 
QQ'  (since  they  are  homologous  sides  of  the  equal  triangles 
NQQ',  PRP").  Hence  RRI  =  PP'  +  QQ')  and  therefore 


Area  MNR  =  area  MNP  +  area  MNQ  ; 

which  proves  the  proposition. 

If  the  origin  of  moments  is  so  taken  that  the  moments  of 
AB  and  BC  have  opposite  signs,  the  demonstration  needs  modi- 
fication. The  student  should  attempt  the  proof  of  this  case  for 

i  •  ir  VT>T,N      v,V»o~    fJM\^-     |\J  I\A  ^   -    /viMGL 

himself. 

^^     •••- 


32 


GRAPHIC    STATICS.  ^  , 

"    0 


49.  Moment  of  a  Couple.  —  Definition.  —  The  moment  of  a 
_  -N  couple  about  any  point  in  its  plane  taken  as  an  origin  is  the 

algebraic  sum  of  the  moments  of  the  two  forces  composing  it 
with  reference  to  the  same  origin. 

Proposition.  —  The  moment  of  a  couple  is  the  same  for  every 
origin  in  its  plane,  and  is  numerically  equal  to  the  product  of 
the  magnitude  of  either  force  into  the  arm  of  the  couple. 
(Art.  7.) 

The  proof  of  this  proposition  can  readily  be  supplied  by  the 
student. 

50.  Moment  of  the  Resultant  of  Any  System.  —  Proposition. 
-  The  algebraic  sum  of  the  moments  of  any  given  complanar 

forces,  with  reference  to  any  origin  in  their  plane,  is  equal  to 
the  moment  of  their  resultant  force  or  resultant  couple  with 
reference  to  the  same  origin. 

The  construction  employed  in  proving  this  proposition  is 
similar  to  that  used  in  Art.  27,  which  the  student  may  profit- 
ably review  at  this  point.  Referring  to  Fig.  9,  let  the  given 
forces  be  represented  in  magnitude  and  direction  by  AB,  BC, 
CD,  DE,  and  in  lines  of  action  by  ab,  be,  cd,  de.  Let  the 
origin  of  moments  be  any  point  in  the  space  diagram.  As  in 
Art.  27,  replace  AB  by  AO,  OB,  acting  in  lines  ao,  ob  ;  replace 
BC  by  BO,  OC,  acting  in  lines  bo,  oc  ;  replace  CD  by  CO,  OD, 
acting  in  lines  co,  od;  and  replace  DE  by  DO,  OE,  acting  in 
lines  do,  oe.  (For  brevity,  we  refer  to  a  force  as  AB,  meaning 
"the  force  represented  in  magnitude  and  direction  by  AB."} 

Now  by  Art.  48,  we  have,  whatever  the  origin, 

Moment  of  A  B  =  moment  of  A  O  +  moment  of  OB; 
«    BC=       "         "   BO+        "        "    OC; 
"         "    CD  =       "         "    CO  +        "       "    OD; 
"         "  DE=       "         "  DO+        "       "    OE. 

Since  the  forces  represented  by  BO  and  OB  have  the  same 
line  of  action,  their  moments  are  numerically  equal  but  have 


MOMENTS   OF   FORCES   AND   OF   COUPLES. 


33 


opposite  signs ;  and  similar  statements  are  true  of  CO  and 
OC,  and  of  DO  and  OD.  Hence,  the  addition  of  the  above 
four  equations  shows  that  the  sum  of  the  moments  of  AI3, 
BC,  CD,  DE,  is  equal  to  the  sum  of  the  moments  of  AO 
and  OE.  Now  the  given  system  has  either  a  resultant  force 
or  a  resultant  couple.  In  the  first  case  the  resultant  of  the 
system  is  the  resultant  of  AO  and  OE,  and  its  moment  is 
equal  to  the  algebraic  sum  of  their  moments,  by  Art.  48. 
In  the  second  case  (which  occurs  only  when  E  coincides  with 
A),  the  resultant  couple  is  composed  ot  AO  and  OE  (which 
in  this  case  are  equal  and  opposite  forces),  and  the  moment 
of  the  couple  is,  by  definition,  equal  to  the  algebraic  sum  of 
the  moments  of  AO  and  OE.  Hence,  in  either  case,  the 
proposition  is  true. 

It  should  be  noticed  that  the  proof  here  given  applies  to 
systems  of  parallel  forces,  as  well  as  to  non-parallel  systems. 
The  proposition  of  Art.  48  could  be  extended  to  the  case  of 
any  number  of  forces,  by  considering  first  the  resultant  of 
two  forces,  then  combining  this  resultant  with  the  third  force, 
and  so  on  ;  but  the  method  would  fail  if,  in  the  process,  it 
became  necessary  to  combine  parallel  forces.  The  method 
here  adopted  is  not  subject  to  this  failure. 

51.  Condition  of  Equilibrium.  —  It  follows  from  what  has 
preceded,  that  if  a  given  system  is  in  equilibrium,  the  alge- 
braic sum  of  the  moments  must  be  zero,  whatever  the  origin. 
For,  in  case  of  equilibrium,  AO  and  OE  (Fig.  9)  are  equal  and 
opposite  and  have  the  same  line  of  action  ;  hence,  the  sum 
of  their  moments  (which  is  the  same  as  the  sum  of  the 
moments  of  the  given  forces)  is  equal  to  zero.  Conversely, 
If  the  algebraic  sum  of  the  moments  is  zero  for  every  origin, 
the  system  must  be  in  equilibrium.  For,  if  it  is  not,  there 
is  either  a  resultant  force  or  a  resultant  couple.  But  the 
moment  of  a  force  is  not  zero  for  any  origin  not  on  its  line 
of  action  ;  and  the  moment  of  a  couple  is  not  zero  for  any 


34 


GRAPHIC   STATICS. 


origin.     For  a  fuller  discussion  of  the  conditions  of  equilibrium, 
see  Arts.  58  and  59. 

52.  Equivalent  Couples. — Proposition.  —  If  a  system  has  for 
its    resultant   a  couple,   it   is   equivalent   to  any   couple  whose 
moment   is   equal  to   the   sum   of  the  moments   of    the  forces 
of  the  system. 

For,  as  already  seen  (Art.  31),  when  the  resultant  is  a 
couple,  the  force  polygon  is  closed.  Let  the  initial  and  final 
points  of  the  force  polygon  coincide  at  some  point  A,  and 
let  O  be  the  pole.  Then  the  forces  of  the  resultant  couple 
are  represented  in  magnitude  and  direction  by  AO,  OA.  Since 
the  position  of  O  is  arbitrary,  the  force  AO  (or  OA)  may  be 
made  anything  whatever  in  magnitude  and  direction.  Also 
the  line  of  action  of  the  force  AO  may  be  taken  so  as  to  pass 
through  any  chosen  point.  Hence,  the  resultant  couple  may 
have  for  one  of  its  forces  any  force  whatever  in  the  plane  of 
the  given  system  ;  and  the  other  force  will  have  such  a  line 
of  action  that  the  moment  of  the  couple  will  be  equal  to  the 
sum  of  the  moments  of  the  given  forces. 

This  reasoning  is  equally  true  if  the  given  system  is  a 
couple.  Hence,  a  couple  is  equivalent  to  any  other  couple 
having  the  same  moment.  In  other  words,  all  couples  wJiosc 
moments  are  equal  are  equivalent ;  and  conversely,  all  equiva- 
lent couples  have  eqtial  moments. 

[NOTE.  — The  construction  above  discussed  fails  if  all  forces  of  the  given  system 
are  parallel  to  the  direction  chosen  for  the  forces  of  the  resultant  couple.  For  then 
the  force  polygon  is  a  straight  line,  and  if  the  pole  is  chosen  in  that  line,  the 
strings  of  the  funicular  polygon  are  parallel  to  the  lines  of  action  of  the  forces,  and 
the  polygon  cannot  be  drawn.  But  in  this  case,  the  system  may  first  be  reduced  to  a 
couple  whose  forces  have  some  other  direction,  and  this  couple  may  be  reduced  to 
one  whose  forces  have  the  direction  first  chosen.  Hence,  the  proposition  stated 
holds  in  all  cases.] 

53.  Moment  of  a  System.  —  Definition.  — The  moment  of  a 
system  of  forces  is  the  algebraic  sum  of  the  moments  of  the 
forces  of  the  system. 


GRAPHIC   DETERMINATION    OF   MOMENTS.  35 

54.  Moments   of   Equivalent    Systems.  —  Proposition.  —  The 
moments  of  any  two   equivalent   systems   of  complanar  forces 
with  respect  to  the  same  origin  are  equal. 

This  follows  immediately  from  the  preceding  articles.  For, 
if  the  two  systems  are  equivalent,  each  is  equivalent  to  the 
same  resultant  force  or  resultant  couple,  and  the  moments  of 
the  two  systems  are  therefore  each  equal  to  the  moment  of 
this  resultant  and  hence  to  each  other. 

§   5  .    Graphic  Determination  of  Moments. 

55.  Proposition.  —  If,  through    any  point   in   the    space  dia- 
gram a  line  be  drawn  parallel  to  a  given  force,  the  distance 
intercepted  upon  it  by  the  two  strings  corresponding  to  that 
force,   multiplied  by  the  pole  distance  of   the  force,  is   equal 
to  the  moment  of  the  force  with  respect  to  the  given  point. 

By  the  strings  "corresponding  to"  a  given  force  are  meant 
the  two  strings  which  intersect  at  a  point  on  its  line  of  action. 

Let  AB  (Fig.  18)  represent  the  magnitude  and  direction  of 
a  force  whose  line  of  action  is  abt  and  let  M  be  the  origin 
of  moments.     Let  O  be  the  pole, 
and  OK  (=H)  the  pole  distance 
of  the  given  force.     Draw  the 
strings  oa,   ob,  and    through   M 
draw  a  line  parallel  to  aby  inter- 
secting oa  and  ob  in  P  and  Q.  FIS.  is 

Then  it  is  to  be  proved  that  the 

moment    of   the   given   force   with    respect   to  M  is  equal  to 
ffxPQ* 

Let  h  equal  the  perpendicular  distance  of  M  from  ab.  Then 
the  required  moment  is  AB  x  //.  But  since  the  similar  triangles 
OAB  and  RPQ  have  bases  AB,  PQ,  and  altitudes  H,  //,  respec- 
tively, it  follows  that 


=~  ;  hence, 
which  proves  the  proposition. 


36  GRAPHIC    STATICS. 

It  should  be  noticed  that  PQ  represents  a  length,  while  H 
represents  a.  force  magnitude.  Hence,  the  moment  of  the  given 
force  with  respect  to  M  is  equal  to  the  moment  of  a  force  H 
with  an  arm  PQ.  [It  may,  in  fact,  be  shown  that  the  given 
force  is  equivalent  to  an  equal  force  acting  in  the  line  PQ 
(whose  moment  about  M  is  therefore  zero),  and  a  couple  with 
forces  of  magnitude  H,  and  arm  PQ.  For  AB  acting  in  ab  is 
equivalent  to  AO  and  OB  acting  in  ao  and  ob  respectively. 
Also,  AO  acting  in  ao  is  equivalent  to  forces  represented  by 
AK  and  KO  acting  respectively  in  PQ  and  in  a  line  through 
P  at  right  angles  to  PQ  ;  and  OB  may  be  replaced  by  forces 
represented  by  OK  and  KB,  the  former  acting  in  a  line  through 
Q  perpendicular  to  PQ,  and  the  latter  in  PQ.  But  AK  and 
KB  are  equivalent  to  AB  ;  hence,  the  proposition  is  proved.] 

56.  Moment  of  the  Resultant  of  Several  Forces.  — The  mo- 
ment of  the  resultant  of  any  number  of  consecutive  forces  in 
the  force  and  funicular  polygons  may  be  found  by  a  method 
similar  to  that  just  described.  Thus,  let  Fig.  19  represent  the 

force  polygon  and  the 
funicular  polygon  for  six 
forces,  and  let  it  be  re- 
quired to  find  the  moment 
of  the  resultant  of  the  four 
forces  represented  in  the 
force  polygon  by  BC,  CD, 
DE,  and  EF,  with  respect 
to  any  point  M.  The  re- 
sultant of  the  four  forces 
is  represented  by  BF,  and 

acts  in  a  line  through  the  intersection  of  ob  and  of.  Through 
M  draw  a  line  parallel  to  BF,  intersecting  ob  and  of  in  P  and  Q 
respectively.  Then  PQ  multiplied  by  OK,  the  pole  distance  of 
BF,  gives  the  required  moment.  This  method  does  not  apply 
to  the  determination  of  the  moment  of  the  resultant  of  several 
forces  not  consecutive  in  the  force  polygon. 


SUMMARY    OF   CONDITIONS   OF   EQUILIBRIUM.  37 

57.  Moments  of  Parallel  Forces. — The  method  of  Arts.  55 
and  56  is  especially  useful  when  it  is  desired  to  find  the  mo- 
ments of  any  or  all  of  a  system  of  parallel  forces  ;  since  with 
such  a  system  the  pole  distance  is  the  same  for  all  forces,  and 
the  moments  are  therefore  proportional  to  the  intercepts  found 
by  the  above  method. 

Example.  —  Assume  five  parallel  forces  at  random;  choose 
an  origin,  and  determine  their  separate  moments,  also  the 
moment  of  their  resultant,  by  the  method  of  Arts.  55  and  56. 

§  6.    Summary  of  Conditions  of  Equilibrium. 

58.  Graphical  and  Analytical  Conditions  of  Equilibrium  Com- 
pared.—  It  has   been   shown  (Art.   35)  that   the  conditions  of 
equilibrium  for  a  system  of  complanar  forces  acting  on  a  rigid 
body  are  two  in  number  : 

(I)  The  force  polygon  must  close. 

(II)  Any  funicular  polygon  must  close. 

The  analytical  conditions  *  are  the  following  : 

(1)  The    algebraic   sum  of  the  resolved  parts  of  the  forces 
in  any  direction  must  be  zero. 

(2)  The  algebraic  sum  of  their  moments  for  any  origin  must 
be  zero. 

The  condition  (i)  is  readily  seen  to  be  equivalent  to  (I). 
For  if  the  sides  of  the  force  polygon  be  orthographically 
projected  upon  any  line,  their  projections  will  represent  in 
magnitude  and  direction  the  resolved  parts  of  the  several 
forces  parallel  to  the  line ;  and,  if  the  force  polygon  is  closed, 
the  algebraic  sum  of  these  projections  is  zero,  whatever  the 
direction  of  the  assumed  line.  (See  Art.  22.)  It  may  also 
be  seen  that  condition  (II)  carries  with  it  (2).  For,  if  every 
funicular  polygon  closes,  the  system  is  equivalent  to  two  equal 
and  opposite  forces  having  the  same  line  of  action  (Arts..  31, 

*  See  Minchin's  Statics,  Vol.  I,  p.  114. 


38  GRAPHIC   STATICS. 

35,  and  50) ;  and  the  sum  of  the  moments  of  these  two  forces 
must  be  zero. 

A  further  comparison  may  be  made.  The  analytical  condi- 
tion (2)  carries  (i)  with  it  ;  and  similarly  the  graphical  condi- 
tion (II)  carries  with  it  the  condition  (I).  That  (2)  includes 
(i)  may  be  seen  as  follows  :  If  the  sum  of  the  moments  is  zero 
for  one  origin  M^  there  can  be  no  resultant  couple,  neither  can 
there  be  a  resultant  force  unless  with  a  line  of  action  passing 
through  MI.  If  the  sum  of  the  moments  is  zero  for  two 
origins,  Mi  and  M2,  the  resultant  force,  if  one  exists,  must 
act  in  the  line  M^M*.  If  the  sum  of  the  moments  is  zero 
also  for  a  third  origin  M3)  not  on  the  line  MiM-2,  there  can 
be  no  resultant  force.  It  follows  at  once  that  condition  (i) 
must  hold. 

That  (II)  includes  (I)  may  be  shown  as  follows  :  Let  A  and 
E  be  the  first  and  last  points  of  the  force  polygon,  and  choose 
a  pole  O\.  Then,  if  the  funicular  polygon  closes,  O\A  and 
OiE  are  parallel.  Choose  a  second  pole  O2,  and,  if  the  funic- 
ular polygon  again  closes,  O2A  and  O2E  are  parallel.  A£ 
must  then  be  parallel  to  OiO2,  unless  A  and  E  coincide. 
Now,  choose  a  third  pole  O3,  not  on  O\O^\  if  the  funicular 
polygon  for  this  pole  closes,  O3A  and  OSE  must  be  parallel. 
But  this  is  impossible  unless  A  and  E  coincide ;  that  is,  unless 
condition  (I)  holds. 

The  last  result  may  be  reached  in  another  way.  With  any 
pole  O  draw  a  funicular  polygon  and  suppose  it  to  close.  The 
system  is  thus  reduced  to  two  forces  acting  in  the  same  line 
oa.  Hence,  there  is  no  resultant  couple,  and  if  there  is  a 
resultant  force,  its  line  of  action  is  oa.  With  the  same  pole 
draw  a  second  funicular  polygon,  the  first  side  being  o'a', 
parallel  to  oa.  If  this  polygon  closes,  there  can  be  no  result- 
ant force,  for  if  one  existed  it  would  act  in  the  line  o'a' ;  and 
there  would  thus  be  two  resultant  forces,  acting  in  different 
lines  oa  and  o'a',  which  is  impossible. 


SUMMARY   OF   CONDITIONS   OF    EQUILIBRIUM.  39 

59.  Summary.  —  It  is  now  evident  that  the  conditions  neces- 
sary to  insure  equilibrium  may  be  stated  in  several  different 
ways,  both  analytically  and  graphically.  To  summarize  : 

A.  Analytically  :  There  will  be  equilibrium  if  either  of  the 
following  conditions  is  satisfied  : 

(1)  The  sum  of  the  moments  is  zero  for  each  of  three  points 
not  in  the  same  line. 

(2)  The  sum  of  the  moments  is  zero  for  each  of  two  points, 
and  the  sum  of  the  resolved  parts  is  zero  for  a  line  not  per- 
pendicular to  the  line  joining  those  two  points. 

(3)  The  sum  of  the  moments  is  zero  for  one  point,  and  the 
sum  of  the  resolved  parts  is  zero  for  each  of  two  directions. 

B.  Graphically :  There  will  be  equilibrium  if  either  of  these 
three  conditions  is  satisfied  : 

(1)  A  funicular  polygon  closes  for  each  of  three  poles  not  in 
the  same  line. 

(2)  Two  funicular  polygons  close  for  the  same  pole. 

(3)  One  funicular  polygon  closes  and  the  force  polygon  closes. 


CHAPTER  III.     INTERNAL  FORCES  AND 
STRESSES. 

§  I  .    External  and  Internal  Forces. 

60.  Definitions.  —  It  was  stated  in  Art.  3  that  every  force 
acting  upon  any  body  is  exerted  by  some  other  body.  In  what 
precedes,  we  have  been  concerned  only  with  the  effects  pro- 
duced by  forces  upon  the  bodies  to  which  they  are  applied. 
It  has  therefore  not  been  needful  to  consider  the  bodies  which 
exert  the  forces.  It  is  now  necessary  to  consider  forces  in 
another  aspect. 

The  forces  applied  to  any  particle  of  a  body  may  be  either 
external  or  internal. 

An  external  force  is  one  exerted  upon  the  body  in  question 
by  some  other  body. 

An  internal  force  is  one  exerted  upon  one  portion  of  the  body 
by  another  portion  of  tlie  same  body. 

It  is  important  to  note,  however,  that  the  same  force  may  be 
internal  from  one  point  of  view,  and  external  from  another. 
Thus,  if  a  given  body  be  conceived  as  made  up  of  two  parts, 
X  and  y,  a  force  exerted  upon  X  by  Y  is  internal  as  regards 
the  whole  body,  but  external  as  regards  the  part  X.  Thus,  let 

_  AB   (Fig.    20)    represent 

/  ^  s\ 

-+  -  [  !  \^  --  ^        a  bar,  acted  upon  by  two 

A 

Fig.  20 


forces  of  equal  magnitude 


applied  at  the  ends  paral- 
lel to  the  length  of  the  bar,  in  such  a  way  as  to  tend  to  pull  it 
apart.  These  two  forces  are  exerted  upon  AB  by  some  other 


40 


EXTERNAL   AND    INTERNAL   FORCES.  4I 

bodies  not  specified.  If  the  whole  bar  be  considered,  the 
external  forces  acting  upon  it  are  simply  the  two  forces 
named. 

But  suppose  the  body  under  consideration  is  AC,  a  portion  of 
AB.  The  external  forces  acting  upon  this  body  are  (i)  a  force 
at  A,  already  mentioned,  and  (2)  a  force  at  C,  exerted  upon  AC 
by  CB.  This  latter  force  is  internal  to  the  bar  AB,  but  external 
to  AC. 

61.    Conditions  of  Equilibrium  Apply  to  External  Forces.  —  In 

applying  the  conditions  of  equilibrium  deduced  in  previous 
articles,  it  must  be  remembered  that  only  external  forces  are 
referred  to.  It  is  also  important  to  notice  that  the  principles 
apply  to  any  body  or  any  portion  of  a  body  in  equilibrium  ;  and 
the  system  of  forces  in  every  case  must  include  all  forces  that 
are  external  to  the  body  or  portion  of  a  body  in  question. 

Thus,  if  the  bar  AB  (Fig.  20)  be  in  equilibrium  under  the 
action  of  two  opposite  forces  P  and  Q,  applied  at  A  and  B 
respectively,  as  shown  in  the  figure,  the  principles  of  equilibrium 
may  be  applied  either  to  the  whole  bar,  or  to  any  part  of  it,  as 
AC. 

(a)  For  the  equilibrium  of  the  whole  bar  AB,  we  must  have 
P=Q,  these  being  supposed  the  only  external  forces  acting  on 
the  bar. 

(ft)  For  the  equilibrium  of  AC,  the  force  exerted  upon  AC  by 
CB  must  be  equal  and  opposite  to  P.  This  latter  force  is 
external^®  AC,  though  internal  \&  the  whole  bar. 

The  method  just  illustrated  is  of  frequent  use  in  the  investi- 
gation of  engineering  structures.  It  is  often  desired  to  deter- 
mine the  internal  forces  acting  in  the  members  of  a  structure, 
and  the  general  method  is  this  :  Direct  the  attention  to  such  a 
portion  of  the  whole  structure  or  body  considered  that  the 
internal  forces  which  it  is  desired  to  determine  shall  be  external 
to  the  portion  in  question.  (See  Art.  67.) 


42  GRAPHIC    STATICS. 

§  2.    External  and  Internal  Stresses. 

62.  Newton's  Third  Law.  —  Let  X  and  Y  be  any  two  portions 
of  matter ;  then  if  X  acts  upon  Y  with  a  certain  force,  Y  acts 
upon  X  with  a  force  of  equal  magnitude  in  the  opposite  direc- 
tion.    This  is  the  principle   stated  in  Newton's  third  law  of 
motion, — that  "to  every  action  there  is  an  equal  and  contrary 
reaction."     It  is  justified  by  universal  experience. 

63.  Stress. — Definition.  —  Two  forces  exerted  by  two  por- 
tions of  matter  upon  each  other  in  such  a  way  as  to  constitute 
an  action  and  its  reaction,  make  up  a  stress. 

Illustrations. — The  earth  attracts  the  moon  with  a  certain 
force,  and  the  moon  attracts  the  earth  with  an  equal  and  opposite 
force.  The  two  forces  constitute  a  stress. 

Two  electrified  bodies  attract  (or  repel)  each  other  with  equal 
and  opposite  forces.  These  two  forces  constitute  a  stress. 

Any  two  bodies  in  contact  exert  upon  each  other  equal  and 
opposite  pressures  (forces),  constituting  a  stress. 

By  the  magnitude  of  a  stress  is  meant  the  magnitude  of  either 
of  its  forces. 

64.  External  and  Internal  Stresses.  —  It  has  been  seen  that 
two  portions  of  matter  are  concerned  in  every  stress.     Now  the 
two  portions  may  be  regarded  either  as  separate  bodies,  or  as 
parts  of  a  body  or  system  of  bodies  which  include  both. 

A  stress  acting  between  two  parts  of  the  same  body  (or 
system  of  bodies)  is  an  internal  stress  as  regards  that  body 
or  system. 

A  stress  acting  between  two  distinct  bodies  is  an  external 

O 

stress  as  regards  either  body. 

It  is  important  to  notice  that  the  same  stress  may  be  internal 
from  one  point  of  view,  and  external  from  another.  Thus,  if 
a  given  body  be  considered  as  made  up  of  two  parts  X  and  F, 
a  stress  exerted  between  X  and  F  is  internal  to  the  whole  body, 
but  external  to  either  X  or  F. 


EXTERNAL   AND    INTERNAL   STRESSES.  43 

Illustration.  — Consider  a  body  AB  (Fig.  21)  resting  upon  a 
second  body  Y,  and  supporting  another  body  X,  as  shown.  If 
the  weight  of  the  body  AB  be  disregarded, 
the  forces  acting  upon  it  are  (i)  the  down- 
ward pressure  (say  P)  exerted  by  X  at  the 
surface  A,  and  the  upward  pressure  (say  Q) 
exerted  by  Fat  B  ;  these  forces  being  equal 
and  opposite,  since  the  body  is  in  equilibrium. 
Now  the  body  X  is  acted  upon  by  AB  with 
a  force  equal  and  opposite  to  P,  and  these  Fis- S1 

two  forces  constitute  a  stress  which  is  external  to  AB.  There 
is  also  an  external  stress  exerted  between  AB  and  Y  at  B. 
But  let  AB  be  considered  as  made  up  of  two  parts,  AC  and  CB. 
Then  (Art.  60)  CB  exerts  upon  AC  a  force  upward,  and  AC 
exerts  upon  CB  a  force  downward.  These  two  forces  are  an 
action  and  its  reaction,  and  constitute  a  stress  which  is  internal 
to  the  body  AB.  This  same  stress  is,  however,  external  to 
either  AC  or  CB.  An  equivalent  stress  evidently  exists  at 
every  section  between  A  and  B.  (When  we  refer  to  the  force 
acting  upon  CB  at  C,  we  mean  the  resultant  of  all  forces  exerted 
upon  the  particles  of  CB  by  the  particles  of  AC.  This  resultant 
is  made  up  of  very  many  forces  acting  between  the  particles. 
Also  the  stress  at  C  means  the  stress  made  up  of  the  two 
resultants  of  the  forces  exerted  by  AC  and  £Z?  upon  each  other.) 

65.  Three  Kinds  of  Internal  Stress.  —  It  is  evident  that  the 
internal  stress  at  C  in  the  body  AB  (Fig.  20)  depends  upon 
the  external  forces  applied  to  the  body.  If  the  forces  at  A  and 
B  cease  to  act,  the  forces  exerted  by  AC  and  CB  upon  each 
other  become  zero.  If  the  forces  at  A  and  B  are  reversed  in 
direction,  so  also  are  those  at  C.  (As  a  matter  of  fact,  the 
particles  of  AC  exert  forces  upon  those  of  CB,  even  if  the 
external  forces  do  not  act.  But  if  the  external  forces  applied 
to  CB  are  balanced,  the  resultant  of  the  forces  exerted  on  CB 
by  A  C  is  zero.) 


44  GRAPHIC    STATICS. 

The  nature  of  the  internal  stress  at  any  point  in  a  body  is 
thus  seen  to  depend  upon  the  external  forces  applied  to  the  body. 

Now,  if  we  consider  two  adjacent  portions  of  a  body  (as 
the  parts  X  and  F.  Fig.  22)  separated  by  a  plane  surface,  the 

external  forces  may  have  either  of  three 
tendencies:  (i)  to  pull  X  and  Y  apart 
in  a  direction  perpendicular  to  the  plane 
of  separation  ;  (2)  to  push  them  together 

in  a  similar  direction  ;  (3)  to  slide  each  over  the  other  along  the 
plane  of  separation.  Corresponding  to  these  three  tendencies, 
the  stress  between  X  and  Y  may  be  of  either  of  three  kinds  : 
tensile,  compressive,  or  shearing. 

A  tensile  stress  is  such  as  comes  into  action  to  resist  a 
tendency  of  the  two  portions  of  the  body  to  be  pulled  apart  in 
the  direction  of  the  normal  to  their  surface  of  separation. 

A  compressive  stress  is  such  as  comes  into  action  to  resist  a 
tendency  of  X  and  Y  to  move  toward  each  other  along  the 
normal  to  the  surface. 

A  shearing  stress  is  such  as  acts  to  resist  a  tendency  of  X 
and  Fto  slide  over  each  other  along  the  surface  between  them. 

In  case  of  a  tensile  stress,  the  force  exerted  by  X  upon  Y 
has  the  direction  from  Y  toward  X;  and  the  force  exerted  by 
Fupon  X  has  the  direction  from  X  toward  F. 

In  case  of  a  compressive  stress,  the  force  exerted  by  X  upon 
Fhas  the  direction  from  X  toward  F;  and  the  force  exerted  by 
Fupon  X  has  the  direction  from  F  toward  X. 

In  the  case  of  a  shearing  stress,  the  force  exerted  by  X 
upon  Fmay  have  any  direction  in  the  plane  of  separation  ;  the 
force  exerted  by  Fupon  X  having  the  opposite  direction. 

If  X  and  F  are  separate  bodies,  instead  of  parts  of  one 
body,  a  similar  classification  may  be  made  of  the  kinds  of 
stress  between  them  ;  but  with  these  we  shall  have  no  occasion 
to  deal.  The  terms  tensile  stress,  compressive  stress,  and  shear- 
ing stress  (or  tension,  compression,  and  shear)  are  usually  applied 
only  to  internal  stresses. 


EXTERNAL   AND   INTERNAL   STRESSES. 


45 


66.  Strain.  —  In  what  has  preceded,  the  bodies  dealt  with 
have  been  regarded  as  rigid ;  that  is,  the  relative  positions  of 
the  particles  of  any  body  have  been  regarded  as  remaining 
unchanged.  But,  as  remarked  heretofore,  no  known  body  is 
perfectly  rigid.  If  no  external  forces  act  upon  a  body,  its 
particles  take  certain  positions  relative  to  each  other,  and  the 
body  has  what  is  called  its  natural  shape  and  size.  If  external 
forces  are  applied,  the  shape  and  size  will  generally  be  changed  ; 
the  body  is  then  said  to  be  in  a  state  of  strain.  The  deforma- 
tion produced  by  any  system  of  applied  forces  is  called  the 
strain  due  to  those  forces.  The  nature  of  this  strain  in  any 
case  depends  upon  the  way  in  which  the  forces  are  applied.  It 
is  unnecessary  to  treat  this  subject  further  at  this  point,  since 
we  shall  at  present  be  concerned  only  with  problems  in  the 
treatment  of  which  it  will  be  sufficiently  correct  to  regard  the 
bodies  as  rigid. 

[NOTE.  —  There  is  a  lack  of  uniformity  among  writers  in  regard  to  the  meanings 
attached  to  the  words  stress  and  strain.  It  may,  therefore,  be  well  to  explain  again 
at  this  point  the  way  in  which  these  words  are  used  in  the  following  pages.  The 
word  stress  should  be  employed  only  in  the  sense  above  defined,  as  consisting  of  two 
equal  and  opposite  forces  constituting  an  action  and  its  reaction.  The  two  forces  are 
exerted  respectively  by  two  bodies  or  portions  of  matter  upon  each  other.  An 
internal  stress  is  a  stress  between  two  parts  of  the  same  body.  An  internal  force  is 
one  of  the  forces  of  an  internal  stress.  It  is  intended  in  what  follows  to  use  the 
word?  "  internal  stress  "  (or  simply  "  stress  r')  only  when  both  the  constituent  forces 
are  referred  to;  and  when  only  one  of  the  forces  is  meant,  to  use  the  words  "inter- 
nal force"  (or  simply  "force").  It  will  be  noticed,  therefore,  that  in  the  following 
pages  the  words  "force  in  a  bar  "  are  frequently  used  where  many  writers  would  say 
"  stress."  This  departure  from  the  usage  of  many  high  authorities  seems  justified  by 
the  following  considerations :  (i)  It  agrees  with  the  usage  which  is  being  adopted 
by  the  highest  authorities  in  pure  mechanics.  (2)  It  is  desirable  that  the  nomencla- 
ture of  technical  mechanics  shall  agree  with  that  of  pure  mechanics,  so  far  as  they 
deal  with  the  same  conceptions.  The  definition  of  strain  above  given  is  in  conform- 
ity with  the  usage  of  the  majority  of  the  more  recent  text-books.  But  it  is  not  rare 
to  find  in  technical  literature  the  word  strain  used  in  the  sense  of  internal  stress  as 
above  defined.  Such  use  of  the  word  should  be  avoided.] 


I. 

« 


46  GRAPHIC   STATICS. 

§  3.    Determination  of  Internal  Stresses. 

67.  General  Method.  —  The  stresses  exerted  between  the 
parts  of  a  body  may  or  may  not  be  completely  determinate  by 
means  of  the  principles  already  deduced.  But  in  all  cases 
these  principles  suffice  for  their  partial  determination.  The 
general  method  employed  is  always  the  same,  and  will  now  be 
illustrated.  As  heretofore  we  deal  only  with  complanar  forces. 
Let  XY  (Fig.  23)  represent  a  body  in  equilibrium  under  the 
action  of  any  known  external  forces  as  shown.  Now  conceive 
the  body  to  be  divided  into  two  parts  as  X  and  Y,  separated 
by  any  surface.  The  particles  of  X  near  the  surface  exert 
upon  those  of  Y  certain  forces,  and  are  in  return  acted 

upon  by  forces  exerted  by 
the  particles  of    Y.    These 

forces   are  internal    as  re- 
p 

\- — ^  gards  the  whole  body.     In 

P«  order  to  determine  them  so 

•  3  Fig.  23 

far  as  possible,  we  proceed 

as  follows  :  Let  the  resultant  of  all  the  forces  exerted  by  Y 
upon  Xb&  called  T\  then  T  is  either  a  single  force  or  a  couple. 
Now  apply  the  conditions  of  equilibrium  to  the  body  X.  The 
external  forces  acting  on  X  are  Plt  P*,  P3,  and  7!  Since  Plt  P.>, 
and  P3  are  supposed  known,  T  can  be  determined.  In  fact, 
T  is  equal  and  opposite  to  the  resultant  of  Ply  P.,,  and  P3. 

So  much  can  always  be  determined.  But  T  is  the  resultant 
of  a  great  number  of  forces  acting  on  the  various  particles 
of  X\  and  these  separate  forces  cannot  in  general  be  deter- 
mined by  methods  which  lie  within  the  scope  of  this  work. 

The  general  principle  just  illustrated  may  be  stated  as 
follows  : 

If  a  body  in  equilibrium  under  any  external  forces  be  conceived 
as  made  up  of  tivo  parts  X  and  Y,  then  the  internal  forces 
exerted  by  X  upon  Y,  together  witJi  the  external  forces  acting  on 
Y,  form  a  system  in  equilibrium. 


DETERMINATION    OF    INTERNAL   STRESSES. 


47 


As  an  immediate  consequence,  we  may  state  that  the  result- 
ant of  the  forces  exerted  by  X  upon  Y  is  equivalent  to  the  result- 
ant of  the  external  forces  acting  on  X ;  and  is  eqnal  and  opposite 
to  the  resultant  of  the  external  forces  acting  upon  Y. 

Example.  —  Assume  a  bar  of  known  dimensions,  and  the 
magnitudes,  directions,  and  points  of  application  of  five  forces 
acting  on  it.  Then  (i)  determine  a  sixth  force  which  will 
produce  equilibrium  ;  and  (2)  assume  the  bar  divided  into  two 
parts  and  find  the  resultant  of  the  forces  exerted  by  each  part 
on  the  other. 

68.  Jointed  Frame.  —  In  certain  ideal  cases  (corresponding 
more  or  less  closely  to  actual  cases),  the  internal  forces  may  be 
more  completely  determined.  The  most  important  of  these 
cases  is  that  which  will  be  now  considered.  Conceive  a  rigid 
body  made  up  of  straight  rigid  bars  hinged  together  at  the  ends. 
Assume  the  following  conditions  : 

(1)  The  hinges  are  without  friction. 

(2)  All   external  forces  acting  on   the  body  are  applied  at 
points  where  the  bars  are  joined  together. 

The  meaning  of  these  conditions  will  be  seen  by  reference  to 
Fig.  24.  The  three  bars  X,  Y,  and  Z  are  connected  by  a  "  pin 
joint,"  the  end  of  each  bar  having  a  hole 
or  "eye"  into  which  is  fitted  a  pin. 
(Of  course  the  three  bars  cannot  be  in 
the  same  plane,  but  they  may  be  nearly 
so,  and  will  be  so  assumed  in  what  fol- 
lows.) Condition  (i)  is  satisfied  if  the 
pin  is  assumed  frictionless.  The  effect  of  this  is  that  the  force 
exerted  upon  the  pin  by  any  bar  (and  the  equal  and  opposite 
reaction  exerted  upon  the  bar  by  the  pin)  acts  in  the  normal 
to  the  surfaces  of  these  bodies  at  the  point  of  contact ;  and, 
therefore,  through  the  centers  of  the  pin  and  the  hole.  Condi- 
tion (2)  means  that  any  external  force  (that  is.,  external  to  the 


48 


GRAPHIC    STATICS. 


whole  body)  applied  to  any  bar  is  applied  to  the  end  and  in  a 
line  through  the  center  of  the  pin. 

With  the  connection  as  shown,  the  bars  do  not  exert  forces 
upon  each  other  directly.  But  each  exerts  a  force  upon  the 
pin,  and  any  force  exerted  by  Y  or  Z  upon  the  pin  causes  an 
equal  force  to  be  exerted  upon  X.  (This  is  seen  by  applying 
the  condition  of  equilibrium  to  the  pin.)  Hence,  in  considering 
the  forces  acting  upon  any  bar  as  X,  we  may  disregard  the  pin 
and  assume  that  each  of  the  other  bars  acts  directly  upon  X. 
By  what  has  been  said,  all  such  forces  exerted  upon  X  by  the 
other  bars  meeting  it  at  the  joint  may  be  regarded  as  acting  at 
the  same  point  —  the  center  of  the  pin.  We  therefore  treat  the 
bars  as  mere  "  material  lines,"  and  regard  all  forces  exerted  on 
any  bar  (whether  by  the  other  bars  or  by  outside  bodies)  as 
applied  at  the  ends  of  this  "material  line." 

Since,  with  these  assumptions,  all  forces  acting  on  any  bar 
MN  (Fig.  25)  are  applied  either  at  M  or  Ar,  the  forces  applied 
at  M  must  balance  those  applied  at  N.  The  resultants  of  the 
two  sets  must  therefore  be  equal  and  opposite,  and  have  the 
same  line  of  action  —  namely  MN.  Further,  it  follows  that 
the  stress  in  the  bar,  acting  across  any  plane  perpendicular  to 
its  length,  is  a  direct  tension  or  compression. 

69.  Internal  Stresses  in  a  Jointed  Frame.  —  Let  Fig.  25  rep- 
resent a  jointed  frame  such  as  above  described,  in  equilibrium 

under  any  known  external  forces. 
Let  us  apply  the  general  method 
of  Art.  67  to  this  case.  Divide 
the  body  into  two  parts,  X  and  Y, 
by  the  surface  AB  as  shown.  Now 
apply  the  conditions  of  equilibrium 
to  the  body  X.  The  system  of 
forces  acting  upon  this  body  consists  of  Pif  P2t  P&,  and  the  forces 
exerted  by  Y  upon  X  in  the  three  members  cut  by  the  surface 
AB.  By  Art.  68  the  lines  of  action  of  these  forces  are  known, 


DETERMINATION    OF   INTERNAL   STRESSES. 


49 


being  coincident  with  the  axes  of  the  members  cut.  Hence, 
the  system  in  equilibrium  consists  of  six  forces,  three  com- 
pletely known,  and  three  known  only  in  lines  of  action. 
The  determination  of  the  unknown  forces  in  magnitude  and 
direction  is  then  a  case  under  the  general  problem  discussed  in 
Art.  40. 

Nature  of  the  stresses.  —  As  soon  as  the  direction  of  the 
force  acting  upon  X  in  any  one  of  the  members  cut  is  known, 
the  nature  of  the  stress  in  that  member  (whether  tension  or 
compression)  is  known.  For  a  force  toivard  X  denotes  com- 
pression ;  while  a  force  aivay  from  X  denotes  tension.  (Art. 

65.) 

If  a  section  can  be  taken  cutting  only  two  members,  the 
forces  in  these  may  be  found  by  the  force  polygon  alone.  The 
same  is  true,  if  any  number  of  members  are  cut,  but  the  stresses 
in  all  but  two  are  known. 

The  methods  described  in  the  last  three  articles  will  find 
frequent  application  in  the  chapters  on  roof  and  bridge  trusses, 
Part  II. 

Example.  — Assume  a  jointed  frame  similar  to  the  one  shown 
in  Fig.  25,  and  let  external  forces  act  at  all  the  joints.  Then 
(i)  assume  all  but  three  of  the  forces  known  in  magnitude  and 
direction  and  determine  the  remaining  three  so  as  to  produce 
equilibrium.  (2)  Take  a  section  cutting  three  members  and 
determine  the  stresses  in  those  members. 

70.  Indeterminate  Cases.  —  If,  in  dividing  the  frame,  more 
than  three  members  are  cut,  the  number  of  unknown  forces  is 
too  great  to  admit  of  the  determination  of  their  magnitudes. 
In  such  a  case,  it  may  happen  that  a  section  elsewhere  through 
the  body  will  cut  but  three  members  ;  and  that  after  the  deter- 
mination of  the  stresses  in  these  three,  another  section  can  be 
taken  cutting  but  three  members  whose  stresses  are  unknown. 
So  long  as  this  can  be  continued,  the  determination  of  the 


50  GRAPHIC    STATICS. 

internal  stresses  can  proceed.  Thus,  in  Fig.  26,  if  a  section 
be  first  taken  at  AB,  there  are  four  unknown  forces  to  be 
determined.  But,  if  the  section  A'B'  be  first  taken,  the 
stresses  in  the  three  members  cut  may  be  determined ;  after 

which   the   section   AB   will 
\Ar  introduce  but  three  unknown 

stresses. 

There   may,   however,    be 

cases  in  which  the  stresses 
iri"  s^j  cannot  all  be  determined  by 

any  method.  With  such  ac- 
tually indeterminate  cases  we  shall  not  usually  have  to  deal.  It 
should  be  noticed,  also,  that  even  when  only  three  members 
are  cut,  the  problem  is  indeterminate  if  these  three  intersect  in 
a  point.  As  in  the  case  just  discussed,  this  indeterminateness 
may  be  either  actual  or  only  apparent ;  in  the  latter  case  it  may 
be  treated  as  above  indicated. 

No  attempt  is  here  made  to  develop  all  methods  that  are 
applicable  or  useful  in  the  determination  of  stresses  in  jointed 
frames.  Some  of  these  are  best  explained  in  connection  with 
the  actual  problems  giving  rise  to  them.  We  have  sought  here 
only  to  explain  and  clearly  illustrate  general  principles. 

71.    Funicular   Polygon  Considered  as  Jointed  Frame.  —  Let 

ab,  be,  cd,  da  (Fig.  27)  be  the  lines  of  action  of  four  forces  in 
equilibrium,  the  force  polygon  being  ABCDA.  Choosing  a 


pole,  draw  any  funicular  polygon,  as  the  one  shown.     Now  let 
the  body  upon  which  the  forces  act  be  replaced  by  a  jointed 


DETERMINATION   OF    INTERNAL    STRESSES.  5I 

frame  whose  bars  coincide  with  the  sides  of  the  funicular 
polygon.  If  at  the  joints  of  this  frame  the  given  forces  be 
applied,  the  frame  will  be  in  equilibrium  ;  and  each  bar  will 
sustain  a  tension  or  compression  whose  magnitude  is  repre- 
sented by  the  corresponding  ray  of  the  force  diagram. 

To  prove  this,  we  apply  the  "general  method"  of  Art.  67. 
Consider  any  joint  (as  the  intersection  of  oa  and  ob),  and  let 
the  frame  be  divided  by  a  plane  cutting  these  two  members. 
Then  the  portion  of  the  frame  about  the  joint  is  acted  upon 
by  three  forces  :  AB,  acting  in  the  line  abt  and  forces  acting  in 
the  bars  cut,  their  lines  of  action  being  oa,  ob.  If  the  bar 
oa  sustains  a  compression  and  ob  a  tension,  their  magnitudes 
being  represented  by  OA  and  BO  respectively,  the  portion 
of  the  frame  about  the  joint  will  be  in  equilibrium.  Hence, 
the  tendency  of  the  force  AB  is  to  produce  the  stresses  men- 
tioned in  the  bars  oa,  ob.  In  the  same  way  it  may  be  shown 
that  the  tendency  of  the  force  BC  is  to  produce  in  ob  and 
oc  tensile  stresses  of  magnitudes  OB  and  CO,  respectively. 
Applying  the  same  reasoning  to  each  joint,  it  is  seen  that  every 
part  of  the  frame  will  be  in  equilibrium  if  the  bars  sustain 
stresses  as  follows  :  The  bar  oa  must  sustain  a  compression 
OA  ;  ob  a  tension  OB]  oc  a  tension  OC\  and  od &  compression 
OD.  Hence,  if  the  bars  are  able  to  sustain  these  stresses,  the 
frame  will  be  in  equilibrium. 

If  the  stress  in  any  member  of  the  frame  is  a  tension,  that 
member  may  be  replaced  by  a  flexible  string.  This  is  the 
origin  of  the  name  string  as  applied  to  the  sides  of  the  funic- 
ular polygon.  This  name  is  retained  for  convenience,  but,  as 
just  shown,  it  is  not  always  appropriate. 

72.  Outline  of  Subject. — The  foregoing  pages,  embracing 
Part  I,  have  been  devoted  to  a  development  of  the  principles 
of  pure  statics.  We  pass  next  to  the  application  of  these 
principles  to  special  classes  of  problems. 

Part  II  treats  of   the   determination  of  internal  stresses  in 


52  GRAPHIC   STATICS. 

engineering  structures.  Only  "  simple  "  structures  are  consid- 
ered,—  that  is,  those  whose  discussion  does  not  involve  the 
theory  of  elasticity.  The  structures  considered  include  roof 
trusses,  beams,  and  bridge  trusses. 

Part  III  develops  the  graphic  methods  of  determining  cen- 
troids  (centers  of  gravity)  and  moments  of  inertia  of  plane 
areas,  including  a  short  discussion  of  "  inertia-curves." 


.       ry>  -     : 


PART  II. 
STX£SS£S  IN  SIMPLE  STRUCTURES. 

CHAPTER    IV.     INTRODUCTORY. 

§   i.    Outline  of  Principles  and  MetJiods. 

73.  The  Problem  of  Design.  —  When  any  structure  is  sub- 
jected to  the  action  of  external  forces,  there  are  brought  into 
action  certain  internal  stresses  in  the  several  parts  of  the  struc- 
ture.    The  nature  and  magnitudes  of  these  stresses  depend  upon 
the  external  forces  acting  (Art.  65).     In  designing  the  structure, 
each  part  must  be  so  proportioned  that  the  stresses  induced  in 
it  will  not  become  such  as  to  break  or  injure  the  material. 

To  determine  these  internal  stresses,  when  the  external  forces 
are  wholly  or  partly  given,  is  the  problem  of  design,  so  far  as 
it  will  be  here  treated. 

74.  External  Forces. — The  external  forces  acting  on  a  struc- 
ture must  generally  be  completely  known  before  the  internal 
stresses  can  be  determined.     These  external  forces  are  usually 
only  partly  given,  and  the  first  thing  necessary  is  to  determine 
them  fully. 

The  external  forces  include  (i)  the  loads  which  the  structure 
is  built  to  sustain,  and  (2)  the  reactions  exerted  by  other  bodies 
upon  the  structure  at  the  points  where  it  is  supported.  The 
former  are  known  or  assumed  at  the  outset  and  the  latter  are 
to  be  determined. 

53 


54  GRAPHIC   STATICS. 

75.  Two  Classes  of  Structures.  —  Structures  may  be  divided 
into  two  classes,  according  as  they  may  or  may  not  be  treated 
as  rigid  bodies  in  determining  the  reactions.  This  may  be 
illustrated  as  follows  : 

Let  a  bar  AB  (Fig.  28)  be  supported  in  a  horizontal  position 

at  two  points  A  and  B,  the  supports  being  smooth  so  that  the 

ip  pressures  on  the  bar   at 

£ i—  A    and    B    are    vertical. 

^WM  Let  a  known  load  P  be 
applied  to  the  beam  at  a 
given  point,  and  let  it  be  required  to  determine  the  reactions  at 
A. and  B. 

This  is  a  determinate  problem ;  for  there  are  three  parallel 
forces  in  equilibrium,  two  being  known  only  in  line  of  action. 
This  problem  was  solved  in  Art.  38. 

But  let  AC  (Fig.  29)  be  a  rigid  bar  supported  at  three  points, 
At  B,  and    C,   the    reactions   at   those   points    being   vertical. 
Ip      ,p  ,p  Let  any  known  loads 
^ — L          B                      c           be   applied   at    given 


points,  and  let  it  be 
required  to  determine 
.  39  the  three  reactions. 

This  problem  is  indeterminate ;  for  any  number  of  sets  of 
values  of  the  three  reactions  may  be  found,  which,  with  the 
applied  loads,  would  produce  equilibrium  if  acting  on  a  rigid 
body.  (See  Art.  40.) 

Since,  however,  an  actual  bar  is  not  a  perfectly  rigid  body, 
such  a  problem  as  the  one  just  stated  is,  in  reality,  determinate. 
But  it  cannot  be  solved  without  making  use  of  the  elastic 
properties  of  the  material  of  which  the  body  is  composed. 

The  two  classes  of  problems  are,  therefore,  the  following : 
(i)  those  in  which  the  reactions  can  be  determined  by  treating 
the  structure  as  a  rigid  body,  and  (2)  those  in  which  the 
determination  of  the  reactions  involves  the  theory  of  elasticity. 
We  shall  at  present  deal  only  with  the  former  class  of  prob- 


OUTLINE   OF   PRINCIPLES   AND   METHODS.  55 

lems.     Structures  coming  under  this  class  will  be  called  simple 
structures. 

76.  Truss.  —  A  truss  is  a  structure  made  up  of  straight  bars 
with  ends  joined  together  in  such  a  manner  that  the  whole  acts 
as  a  single  body.     The  ends  of  the  bars  are,  in  practice,  joined 
in  various  ways  ;  but  in  determining  the  internal  stresses,  the 
connections  are  assumed  to  be  such  that  no  resistance  is  offered 
by  a  joint  to  the  rotation  of  any  member  about  it.     Such  a 
structure  to  be  indeformible  must   be  made  up  of  triangular 
elements  ;   for  more  than   three   bars   hinged   together  in  the 
form    of  a   polygon   cannot   constitute    a   rigid  whole.     If  the 
external  forces  are  applied  to  the  truss  only  at  the  points  where 
the  bars  are  joined,  the  internal  stress  at  any  section  of  a  mem- 
ber will  be  a  simple  tension  or  compression,  directed  parallel  to 
the  length  of  the  bar.     (See  Art.  68.) 

The  most  important  classes  of  trusses  are  roof  trusses  and 
bridge  trusses.  The  methods  used  in  discussing  these  classes 
are,  of  course,  applicable  to  any  framed  structures  under  similar 
conditions. 

77.  Loads  on  a  Truss.  — The  loads  sustained  by  a  truss  may 
be  either  fixed  or  moving.     A  fixed  (or  dead}  load  is  one  whose 
point  of  application  and  direction  remain  constant.     A  moving 
(or  live)  load  is  one  whose  point  of  application  passes  through 
a  series  of  positions.     Fixed  loads  may  be  either  permanent  or 
temporary. 

The  loads  on  a  roof  truss  are  usually  all  fixed,  but  are  of 
various  kinds,  viz.,  the  weight  of  the  truss  itself  and  of  the  roof 
covering,  which  is  a  permanent  load ;  the  weight  of  snow 
lodging  on  the  roof,  and  the  pressure  of  wind,  both  of  which 
are  temporary  loads. 

A  bridge  truss  supports  both  fixed  and  moving  loads.  The 
former  include  the  weight  of  the  truss  itself,  of  the  roadway, 
of  all  lateral  and  auxiliary  bracing  (permanent  loads) ;  and  of 
snow  (a  temporary  load).  The  latter  consist  of  moving  trains 


56  GRAPHIC   STATICS. 

in  the  case  of  railway  bridges,  and  of  teams  or  crowds  of  animals 
or  people  in  the  case  of  highway  bridges. 

78.  Combination  of  Stresses  Due  to  Different  Causes.  —  When 
a  truss  is   subject  to  a  variety   of   external  loads,   it  is  often 
convenient  to  consider  the  effect  of  a  part  of  them  separately. 
If  tensile  and  compressive  stresses    are  distinguished  by  signs 
plus  and  minus,  the  stress  in  any  member  due  to  the  combined 
action  of  any  number  of  loads  is  equal  to  the  algebraic  sum  of 
the  stresses  due  to  the  loads  acting  separately. 

A  proof  of  this  proposition  might  be  given ;  but  it  may  be 
accepted  as  sufficiently  evident  without  formal  demonstration. 

79.  Beams.  —  Another   class    of    bodies    to   be    treated    is 
included  under  the  name  beam. 

A  beam  may  be  defined  as  a  bar  (usually  straight)  resting  on 
supports  and  carrying  loads.  The  loads  and  reactions  are 
commonly  applied  in  a  direction  transverse  to  the  length  of  the 
bar  ;  but  this  is  not  necessarily  the  case. 

The  internal  stresses  in  any  section  of  a  beam  are  less 
sjmple  than  those  in  the  bars  of  an  ideal  jointed  frame  such  as 
a  truss  is  assumed  to  be.  A  discussion  of  beams  is  given  in 
Chap.  VI. 

80.  Summary   of    Principles   Needed.  —  It   will   be   well   to 
summarize  at  this  point  the  main  principles  and  methods  which 
will  be  employed  in  the  discussion  of  the  problems  that  follow. 

The  general  problem  presented  by  any  structure  consists 
of  two  parts  :  (a)  the  determination  of  the  unknown  external 
forces  (or  reactions)  and  (b)  the  determination  of  the  internal 
stresses. 

(a)  In  the  case  of  simple  structures  the  unknown  reactions 
are  usually  two  in  number,  and  the  cases  most  commonly  pre- 
sented are  the  following : 

ist.  — Their  lines  of  action  are  known  and  parallel. 

2nd.  —  The  line  of  action  of  one  and  the  point  of  application 
of  the  other  are  known. 


OUTLINE   OF   PRINCIPLES   AND   METHODS. 


57 


Since  all  the  external  forces  form  a  system  in  equilibrium, 
these  two  cases  fall  under  the  general  problems  discussed  in 
Arts.  38  and  39. 

(b)  In  the  case  of  a  jointed  frame  or  truss,  the  lines  of 
action  of  all  internal  forces  are  known,  since  they  coincide 
with  the  axes  of  the  truss  members.  (Art.  68.)  In  determin- 
ing their  magnitudes  we  may  have  to  deal  with  the  following 
problems  in  equilibrium  : 

ist.  — The  system  in  equilibrium  may  be  completely  known, 
except  the  magnitudes  of  two  forces. 

2nd.  — The  magnitudes  of  three  forces  may  be  unknown. 

The  first  case  may  be  solved  by  simply  making  the  force 
polygon  close.  (See  Art.  35.) 

The  second  case  may  be  solved  by  the  method  of  Art.  40, 
which  consists  in  making  the  force  and  funicular  polygons 
close  ;  or  by  the  method  of  Art.  42 ;  or  by  the  principle  of 
moments  (Art.  51). 

The  student  should  be  thoroughly  familiar  with  the  prob- 
lems and  principles  here  referred  to.  In  the  following  chapters 
we  proceed  to  their  application. 

8 1.  Division  of  the  Subject.  —  The  subject  of  the  design  of 
structures,  so  far  as  here  dealt  with,  will  be  treated  in  three 
divisions.  The  first  relates  to  framed  structures  sustaining 
only  stationary  loads ;  the  second  to  beams  sustaining  both 
fixed  and  moving  loads  ;  the  third  to  framed  structures  sustain- 
ing both  fixed  and  moving  loads.  Among  structures  of  the 
first  class,  the  most  important  are  roof  trusses  ;  hence,  these 
are  chiefly  referred  to  in  the  next  chapter.  For  a  similar 
reason,  the  chapter  devoted  to  the  third  class  of  structures 
refers  principally  to  bridge  trusses.  The  chapter  on  beams 
precedes  that  on  bridge  trusses,  for  the  reason  that  the  methods 
used  in  dealing  with  a  beam  under  moving  loads  form  a  useful 
introduction  to  those  employed  in  treating  certain  classes  of 
truss  problems. 


CHAPTER  V.     ROOF  TRUSSES.  —  FRAMED  STRUC- 
TURES   SUSTAINING   STATIONARY    LOADS. 

§    i.    Loads  on  Roof  Trusses. 

82.  Weights  of  Trusses.  —  Among  the  loads  to  be  sustained 
by  a  roof  truss  is  the  weight  of  the  truss  itself.  Before  the 
structure  is  designed,  its  weight  is  unknown.  But,  since  it  is 
necessary  to  know  the  weight  in  order  that  the  design  may  be 
correctly  made,  the  method  of  procedure  must  be  as  follows  : 

Make  a  preliminary  estimate  of  the  weight,  basing  it  upon 
knowledge  of  similar  structures  ;  or,  in  the  absence  of  such 
knowledge,  upon  the  best  judgment  available.  Then  design 
the  various  truss  members,  compute  their  weight,  and  com- 
pare the  actual  weight  of  the  truss  with  the  assumed  weight. 
If  the  difference  is  so  great  as  to  materially  affect  the  design  of 
the  truss  members,  a  new  estimate  of  weight  must  be  made, 
and  the  computations  repeated  or  revised.  No  more  than  one 
or  two  such  trials  will  usually  be  needed. 

As  a  guide  in  making  the  preliminary  estimate  of  weight, 
the  following  formulas  may  be  used.  They  are  taken  from  Mer- 
riman's  "  Roofs  and  Bridges,"  being  intended  to  represent 
approximately  the  data  for  actual  structures,  as  compiled  by 
Ricker  in  his  ''Construction  of  Trussed  Roofs." 

Let  /—span  in  feet  ;  a  =  distance  between  adjacent  trusses  in 
feet  ;  W=  total  weight  of  one  truss  in  pounds.  Then  for 
wooden  trusses 


and  for  wrought  iron  trusses 


58 


LOADS    ON    ROOF    TRUSSES. 


59 


83.  Weight  of  Roof  Covering.  —  The  weight  of  roof  covering 
can  be  correctly  estimated  beforehand  from  the  known  weights 
of  the  materials.     The  following  data  may  be  employed,  in  the 
absence  of  information  as  to  the  specific  material  to  be  used. 
(See  Merriman's  "  Roofs  and  Bridges,"  p.  4.)     The   numbers 
denote  the  weight  in  pounds  per  square  foot  of  roof  surface. 

Shingling  :  Tin,  I  Ib.  ;  wooden  shingles,  2  to  3  Ibs.  ;  iron,  I 
to  3  Ibs.  ;  slate,  10  Ibs.  ;  tiles,  12  to  25  Ibs. 

Sheathing  :  Boards  I  in.  thick,  3  to  5  Ibs. 

Rafters  :   1.5  to  3  Ibs. 

Purlins  :  Wood,  I  to  3  Ibs.  ;  iron,  2  to  4  Ibs. 

Total  roof  covering,  from  5  to  35  Ibs.  per  square  foot  of  roof 
surface. 

84.  Snow  Loads.  — The  weight  of  snow  that  may  have  to  be 
borne  will  differ  in  different  localities.     For  different  sections 
of  the  United  States  the  following  maybe  used  as  the  maximum 
snow  loads  likely  to  come  upon  roofs. 

Maximum  for  northern  United  States,  30  Ibs.  per  square  foot 
of  horizontal  area  covered. 

For  latitude  of  New  York  or  Chicago,  20  Ibs.  per  square  foot. 

For  central  latitudes  in  the  United  States,  lolbs.  per  square  foot. 

The  above  weights  are  given  in  Merriman's  "  Roofs  and 
Bridges."  They  are  in  excess  of  those  used  by  some  Bridge 
and  Roof  companies. 

85.  Wind  Pressure  Loads. — The  intensity  of  wind  pressure 
against  any  surface  depends  upon  two  elements  :  (a)  the  velocity 
of  the  wind,  and  (b)  the  angle   between  the   surface  and  the 
direction  of  the  wind. 

Theory  indicates  that  the  intensity  of  wind  pressure  upon  a 
surface  perpendicular  to  the  direction  of  the  wind  should  be 
proportional  to  the  square  of  the  velocity  of  the  wind  relative  to 
the  surface.  As  an  approximate  law  this  is  borne  out  by 
experiment.  If  /  denotes  the  pressure  per  unit  area,  and  i> 
the  velocity  of  the  wind,  the  law  is  expressed  by  the  formula 


6o 


GRAPHIC    STATICS. 


p  —  kv*.  Here  k  is  proportional  to  the  density  of  air.  Its 
numerical  value  may  be  taken  as  0.0024,  ^  the  "nits  of  force, 
length,  and  time  are  the  pound,  foot,  and  second  respectively. 

If  the  wind  strikes  a  surface  obliquely,  experiment  shows 
that  the  resulting  pressure  has  a  direction  practically  normal  to 
the  surface.  The  tangential  component  is  inappreciable,  owing 
to  the  very  slight  friction  between  air  and  any  fairly  smooth 
surface.  The  intensity  of  the  normal  pressure  depends  upon 
the  angle  at  which  the  wind  meets  the  surface. 

For  a  given  velocity  of  wind  let/»a  denote  the  normal  pressure 
per  unit  area,  when  the  direction  of  wind  makes  an  angle  a 
with  the  surface,  and  /„  the  pressure  per  unit  area  due  to  the 
same  wind  striking  a  surface  perpendicularly.  Then  the  follow- 
ing formula  *  has  been  given  : 

2  sin  a 


It  will  rarely  be  necessary  to  use  values  of  pn  greater  than 
50  Ibs.  per  square  foot.  The  following  table  gives  values  of 
the  coefficient  of  pn  in  the  above  formula  for  different  values 
of  a.  The  value  of  p,,  may  be  taken  ;'as  from  40  to  50  Ibs.  per 
square  foot. 


a 

2  sin  a 

a 

2  sin  a 

I  -f  sin-  a 

I  +  sin2  a 

0° 

o.oo 

50° 

0.97 

10° 

o-34 

6oc 

0.99 

20° 

0.61 

-70° 

I.OO 

30° 

0.80 

80° 

I.OO 

40° 

0.91 

90° 

I.OO 

*  This  formula  is  given  by  various  writers.  It  is  cited  by  Langley  ("  Experiments  in 
Aerodynamics,"  p.  24),  who  attributes  it  to  Duchemin.  Professor  Langley's  elaborate 
experiments  show  so  close  an  agreement  with  the  formula  that  it  may  be  used  without  hesi- 
tation in  estimating  the  pressure  on  roofs. 


ROOF  TRUSS  WITH  VERTICAL  LOADS.        6 1 

§   2.    Roof  Truss  with    Vertical  Loads. 

86.  Notation.  —  The  method  of  determining  internal  stresses 
in  the  case  of  vertical  loading  will  be  explained  by  reference  to 
the  form  of  truss  shown  in  Fig.  30.     The  method  will  be  seen 
to  be  independent  of  the  particular  form  of  the  truss. 

For  designating  the  truss  members  and  the  lines  of  action  of 
external  forces  a  notation  will  be  employed  similar  to  that  used 
in  previous  chapters.  Let  each  of  the  areas  in  the  truss  dia- 
gram be  marked  with  a  letter  or  other  symbol  as  shown  in 
Fig.  30 ;  then  the  truss  member  or  force-line  separating  any 
two  areas  may  be  designated  by  the  two  symbols  belonging  to 
those  areas.  Thus,  the  lines  of  action  of  the  external  forces 
are  ab,  be,  cdt  etc.,  and  the  truss  members  are  ght  hbt  hi,  etc. 
It  is  to  be  noticed  that  the  lines  representing  the  truss  mem- 
bers represent  also  the  lines  of  action  of  forces, — namely,  the 
internal  forces  in  the  members.  The  joint,  or  point  at  which 
several  members  meet,  may  be  designated  by  naming  all  the 
surrounding  letters.  Thus,  bciht  hijg  are  two  such  points. 

87.  Loads  and  Reactions.  — The  loads  now  considered  are 
assumed  to  be  applied  in  a  vertical  direction,  and  to  act  at  the 
upper  joints  of  the  truss.     This  assumption  as  to  the  points  of 
application  may  in  some  cases  represent  very  nearly  the  facts ; 
in   other  cases   the  loads  will,  in  reality,  be  applied  partly  at 
intermediate  points  on  the  truss  members.     If  the  latter  is  the 
case,  the  load  borne  upon  any  member  is  assumed  to  be  divided 
between  the  two  joints  at  its  ends.     In  this  case  the  member 
will  be  subject  not  only  to  direct  tension  or  compression,  but  to 
bending.     With  the  latter  we  are  not  here  concerned,  although 
it  must  always  be  considered  in  designing  the  member. 

The  ends  of  the  truss  are  supposed  to  be  supported  on  hori- 
zontal surfaces,  and  the  reaction  at  each  point  of  support  is 
assumed  to  have  a  vertical  direction. 

If  the  loading  is  symmetrical  with  reference  to  a  vertical  line 

)    0      1 

V   1   ' 


62  GRAPHIC    STATICS. 

through  the  middle  of  the  truss,  it  is  evident  that  each  reaction 
is  equal  to  half  the  total  load.  If  the  loading  is  not  symmetri- 
cal, the  reactions  cannot  be  determined  so  simply.  They  may, 
however,  be  readily  computed  by  either  graphic  or  algebraic 
methods.  Graphically,  the  problem  is  identical  with  that  solved 
in  Art.  38.  The  truss  is  treated  as  a  rigid  body,  the  external 
forces  acting  upon  it  being  the  loads  and  reactions,  which  form 
a  system  of  parallel  forces  in  equilibrium.  Two  of  these  forces 
(the  reactions)  are  unknown  in  magnitude,  but  known  in  line 
of  action.  The  construction  for  determining  their  magnitudes 
is  as  follows  : 

Draw  the  force  polygon  A  BCD EF  for  the  five  loads ;  choose 
a  pole  Oy  draw  rays  OA,  OB,  OC,  etc.,  and  draw  the  funicular 
polygon  as  shown  in  Fig.  30.  The  two  polygons  are  to  be 
completed  by  including  the  reactions  FG,  GA,  and  both  poly- 
gons must  close.  We  may  draw  first  og,  the  closing  line  of  the 
funicular  polygon,  and  then  the  ray  OG  parallel  to  it,  thus 
determining  the  point  G  in  the  force  polygon.  The  reactions 
are  now  shown  in  magnitude  and  direction  by  FG  and  GA. 

88.  Determination  of  Internal  Stresses.  —  When  the  external 
forces  are  all  known,  the  internal  stresses  may  be  found  very 
readily.  The  only  principle  needed  is,  that  for  any  system  of 
forces  in  equilibrium,  the  force  polygon  must  close.  The  con- 
struction will  now  be  explained. 

Considering  any  joint  of  the  truss  (Fig.  30)  as  ghijg^  fix  the 
attention  upon  the  portion  of  the  truss  bounded  by  the  broken 
line  in  the  figure.  This  portion  is  a  body  in  equilibrium  under 
the  action  of  four  forces  whose  lines  of  action  coincide  with  the 
axes  of  the  four  bars  g/i,  hi,  ij,  jg,  respectively.  These  forces 
are  internal  as  regards  the  truss  as  a  whole,  but  external  to  the 
part  in  question  ;  each  force  being  one  of  the  pair  constituting 
the  internal  stress  at  any  point  of  the  bar.  Such  a  force  acts 
from  the  joint  if  the  stress  in  the  bar  is  a  tension  ;  toivard  it  if 
the  stress  is  a  compression.  (Art.  69.) 


ROOF   TRUSS   WITH    VERTICAL   LOADS.  63 

Since  these  four  forces  form  a  system  in  equilibrium,  their 
force  polygon  must  close.  This  condition  will  enable  us  to  fully 
determine  the  magnitudes  of  the  forces,  provided  all  but  two  are 
known,  since  the  polygon  can  then  be  constructed  as  in  Art.  18. 


1 


We  cannot,  however,  begin  with  the  joint  just  considered, 
since  at  first  the  four  forces  are  all  unknown  in  magnitude. 

If,  however,  we  start  with  the  joint  gab/i,  the  polygon  of 
forces  can  be  at  once  drawn.  For,  reasoning  as  above,  it  is 
seen  that  the  portion  of  the  truss  immediately  surrounding 
this  joint  is  in  equilibrium  under  the  action  of  four  forces : 
the  reaction  in  the  line  ga,  the  load  in  the  line  ab,  and  the 
internal  forces  in  the  lines  bh,  Jig.  Of  these  forces,  two  (the 
reaction  and  the  load)  are  completely  known  ;  and  it  is  neces- 
sary only  to  draw  a  polygon  of  which  two  sides  represent  the 
known  forces  and  the  other  two  sides  are  made  parallel  to  the 
members  bJi,  Jig.  Such  a  polygon  is  shown  in  Fig.  30 ;  and 
BH  and  HG  represent  in  magnitude  and  direction  the  forces 
whose  lines  of  action  are  b/i,  Jig. 


64  GRAPHIC    STATICS. 

Evidently,  BH  and  HG  represent  also  the  magnitudes  (Art. 
63)  of  the  internal  stresses  in  the  two  members  bh  and  Jig. 
The  nature  of  these  stresses  may  be  found  as  follows  :  Since 
the  four  forces  represented  in  the  polygon  GABHG  are  in 
equilibrium,  and  since  GA,  AB  are  the  directions  of  two  of 
them,  the  directions  of  the  other  two  must  be  BH,  HG. 
Hence,  BH  acts  toward  the  joint  and  HG  from  it.  This 
shows  that  the  stress  in  bh  is  a  compression,  while  that  in  hg  is 
a  tension. 

Passing  now  to  the  joint  bciJib,  it  is  seen  that  of  the  four 
forces  whose  lines  of  action  meet  there,  two  are  fully  known, 
namely,  the  load  BC  acting  vertically  downward  and  the  inter- 
nal force  in  hb  acting  toward  the  joint  (since  the  stress  is  com- 
pressive),  while  the  remaining  two  (viz.,  the  internal  forces  in 
ci  and  i/i)  are  unknown  in  magnitude  and  direction.  Since, 
however,  the  unknown  forces  are  but  two  in  number,  the  force 
polygon  can  be  completely  drawn,  and  is  represented  by  the 
quadrilateral  HBCIH.  The  directions  of  the  forces  are  found 
as  in  the  preceding  case,  and  it  is  seen  that  the  bars  ci  and  /// 
both  sustain  compressive  stresses. 

The  process  may  be  continued  by  passing  to  the  remaining 
joints  in  succession,  in  such  order  that  at  each  there  remain  to 
be  determined  not  more  than  two  forces.  The  complete  con- 
struction is  shown  in  Fig.  30. 

It  is  evident  that  the  loads  AB  and  EF  might  have  been 
omitted  without  changing  the  stresses  in  any  of  the  truss 
members.  For  their  omission  would  leave  as  the  complete 
force  polygon  for  external  forces  BCDEGB,  and  the  two 
reactions  would  be  GB  and  EG ;  but  the  force  diagram  would 
be  otherwise  unchanged. 

The  great  convenience  of  the  notation  adopted  is  now  seen.* 


*  This  is  known  as  Bow's  notation.  The  notation  adopted  in  Part  I  involves  the  same 
idea,  but  it  is  not  usually  employed  in  works  on  Graphic  Statics,  though  possessing  very 
evident  advantages.  It  was  suggested  to  the  writer  by  its  use  in  certain  of  Professor 
Eddy's  works. 


ROOF  TRUSS  WITH  VERTICAL  LOADS.        65 

The  line  representing  the  stress  in  any  member  is  designated 
in  the  force  diagram  by  letters  similar  to  those  which  designate 
that  member  in  the  truss  diagram.  The  latter  is  evidently  a 
space  diagram  (Art.  11).  The  force  diagram  is  often  called  a 
stress  diagram,  since  it  shows  the  values  of  the  internal  stresses 
in  the  truss  members. 

89.  Reciprocal  Figures.  —  There  are  always  two  ways  of  com- 
pleting the  force  polygon  when  two  of  the  forces  are  known 
only  in  lines  of  action.     (See  Art.  18.)     Either  way  will  give 
correct    results,  but    unless  a  certain   way  be    chosen,   it  will 
become  necessary  to  repeat  certain  lines  in  the  stress  diagram. 
Thus,  if,  in   Fig.   30,  instead  of  GABHG  we  draw  GABH'G, 
the  lines  GH\  H'B  are  not  in  convenient  positions  for  use  in 
the  other  polygons  of  which  they  ought  to  form   sides.     The 
lettering  of  the  diagrams  will  also  be  complicated.     As  an  aid 
in  drawing  the  lines  in  the  most  advantageous  positions,  it  is 
convenient  to  remember  the  fundamental   property   of  figures 
related  in  such  a  way  as  the  force  and  space  diagrams  shown  in 
Fig-  30. 

Such  figures  are  said  to  be  reciprocal  with  regard  to  each 
other.  The  fundamental  property  of  reciprocal  figures  is  that 
for  every  set  of  lines  intersecting  in  a  point  in  either  figure, 
there  is  in  the  other  a  set  of  lines  respectively  parallel  to  them 
and  forming  a  closed  polygon. 

It  is  also  an  aid  to  remember  that  the  order  of  the  sides  in 
any  closed  polygon  in  the  stress  diagram  is  the  same  as  the 
order  of  the  corresponding  lines  in  the  truss  diagram,  if  taken 
consecutively  around  the  joint.  This  usually  enables  us  at 
once  to  draw  the  sides  of  each  force  polygon  in  the  proper 
order. 

90.  Order  of  External  Forces  in  Force  Polygon.  —  It  will  be 
observed  that  in  the  case  above  considered,  in  constructing  the 
force  polygon  for  the  loads  and  reactions,  these  forces  have 
been  taken  consecutively  in  the  order  in  which  their  points  of 


66 


GRAPHIC   STATICS. 


application  occur  in  the  perimeter  of  the  truss.  This  is  a 
necessary  precaution  in  order  that  the  stress  diagram  and  truss 
diagram  may  be  reciprocal  figures,  so  that  no  line  in  the  former 
need  be  duplicated. 

This  requirement  should  be  especially  noticed  in  such  a  case 
as  that  shown  in  Fig.  31,  in  which  loads  are  applied  at  lower  as 


well  as  at  upper  joints.  If  the  reactions  are  found  by  the 
method  of  Art.  87,  without  modification,  the  force  polygon  will 
not  show  the  external  forces  in  the  proper  order,  since  the 
known  forces  are  not  applied  at  consecutive  joints  of  the  truss. 
A  new  polygon  should  therefore  be  drawn  after  the  reactions 
have  been  determined. 

If  desirable  (as  in  some  cases  it  may  be)  to  make  use  of  a 
funicular  polygon  in  which  the  external  forces  are  taken  con- 
secutively, this  may  be  drawn  after  the  reactions  are  deter- 
mined and  the  new  force  polygon  is  drawn. 

If  a  load  be  applied  at  some  joint  interior  to  the  truss,  as  at 
M  (Fig.  31),  then  in  constructing  the  stress  diagram  it  should 


STRESSES  DUE  TO  WIND  PRESSURE.        67 

be  assumed  to  act  at  N,  where  its  line  of  action  intersects  the 
exterior  member  of  the  truss,  and  the  fictitious  member  MN 
inserted.  The  stresses  in  the  actual  truss  members  will  be 
unaffected  by  this  assumption,  and  such  a  device  is  necessary  in 
order  that  the  stress  diagram  may  be  the  true  reciprocal  of  the 
truss  diagram. 

§   3.    Stresses  Due  to    Wind  Pressure. 

91.  Direction  of  Reactions  Due  to  Wind  Pressure.  —  Since 
the  effective  pressure  of  the  wind  has  the  direction  normal  to 
the  surface  of  the  roof  (Art.  85),  it  has  a  horizontal  component 
which  must  be  resisted  by  the  reactions  at  the  supports. 
These  cannot,  therefore,  act  vertically,  as  in  the  case  when 
the  loading  is  vertical.  Their  actual  directions  will  depend 
upon  the  manner  in  which  the  ends  of  the  truss  are  sup- 
ported. 

If  the  ends  of  the  truss  are  immovable,  the  directions  of  the 
reactions  cannot  be  determined,  since  any  one  of  an  infinite 
number  of  pairs  of  forces  acting  at  the  ends  would  produce 
equilibrium.  (The  same  would  be  true  of  the  reactions  due  to 
vertical  loads.)  In  such  a  case  the  usual  assumption  is  one  of 
the  following:  (i)  the  reactions  are  assumed  parallel  to  the 
loads  ;  (2)  the  resolved  parts  of  the  reactions  in  the  horizontal 
direction  are  assumed  equal. 

In  the  case  of  trusses  of  large  span  it  is  not  unusual  to 
support  one  end  of  the  truss  upon  rollers  so  that  it  is  free  to 
move  horizontally,  the  other  end  being  hinged,  or  otherwise 
arranged  to  prevent  both  horizontal  and  vertical  motion.  This 
allows  for  expansion  and  contraction  with  change  of  tempera- 
ture, as  well  as  for  movements  due  to  the  small  distortions  of 
the  truss  under  loads.  With  this  arrangement  the  reaction  at 
the  end  supported  on  rollers  must  be  vertical ;  and  since  the 
point  of  application  of  the  other  reaction  is  known,  both  can 
be  fully  determined  by  the  method  described  in  Art.  39. 


68  GRAPHIC    STATICS. 

92.  Determination  of  Reactions. — The  methods  of  finding 
reactions  will  now  be  explained  for  the  three  cases  mentioned 
in  the  preceding  article  :  (i)  Assuming  both  reactions  parallel 
to  the  wind  ;  (2)  assuming  the  horizontal  resolved  parts  of  the 
two  reactions  equal ;  and  (3)  assuming  one  reaction  vertical. 

(1)  The  first  case  needs  no  explanation,  since  it  is   identical 
with    that    described    in    Art.    87,  except    that    the    loads    and 
reactions  have  a  direction  normal  to  one  surface  of  the   roof, 
instead  of  being  vertical.     It  is  to  be  noticed  that  this  assump- 
tion cannot  be  made  if  the   roof  surface  is  curved,   since  the 
lines  of  action  of   the  forces  will   not  be  parallel.     But  since 
the  direction  of  the  resultant  of  the  loads  will  be  known  from 
the  force  polygon,  both  reactions  may  be  assumed   to  act  par- 
allel to  this  resultant,  and  the  construction  made  as  before. 

(2)  In  the  second  case,  let  each  reaction  be  replaced  by  two 
forces    acting   at   the    support,   one    horizontal   and   the    other 
vertical.     The  two  horizontal  forces  are  known  as  soon  as  the 
force   polygon    for  the    loads    is  drawn,   and   the    two    vertical 
forces  may  be  found  as  in  the  preceding  case,  since  their  lines 
of  action  are  known. 

In  Fig.  32,  let  ab,  be,  cd  be  the  lines  of  action  of  the  wind 
forces.  Let  the  right  reaction  be  considered  as  made  up  of  a 
horizontal  component  acting  in  de  and  a  vertical  component 
acting  in  ef\  and  let  the  left  reaction  be  replaced  by  a  vertical 
component  acting  in  fg  and  a  horizontal  component  acting  in 
ga.  Draw  the  force  polygon  (or  "  load-line  ")  ABCD.  By  the 
assumption  already  made  GA  and  DE  are  to  be  equal,  and 
their  sum  is  to  equal  the  horizontal  resolved  part  of  AD. 
Through  the  middle  point  of  AD  draw  a  vertical  line ;  its 
intersections  with  horizontal  lines  through  A  and  D  determine 
the  points  G  and  E,  so  that  the  two  forces  GA  and  DE  become 
known.  The  only  remaining  unknown  forces  are  the  vertical 
forces  .ZTFand  FG.  Choose  a  pole  O,  draw  rays  to  the  points 
G,  A,  B,  C,  D,  E,  and  then  the  corresponding  strings.  Through 
the  points  determined  by  the  intersection  of  og  withyjf,  and  oe 


STRESSES    DUE   TO   WIND   PRESSURE. 


69 


with  cf,  draw  the  string  of.  The  corresponding  ray  drawn 
from  O  intersects  EG  in  the  point  F,  thus  determining  EF  and 
FG.  The  reactions  are  now  wholly  known  ;  that  at  the  left 
support  being  DF,  and  that  at  the  right  support  FA. 


"*"-—- SL 

'  b  ~~—~/-' 


(3)  For  the  third  case  the  construction  is  shown  in  Fig.  33 
(A)  and  (/?),  for  the  two  opposite  directions  of  the  wind.  The 
method  is  identical  with  that  employed  in  Art.  39.  Only  one 
point  of  the  line  of  action  of  the  left  reaction  is  known,  hence 
this  is  taken  as  the  point  of  intersection  of  the  corresponding 
strings  of  the  funicular  polygon.  One  of  these  strings  can  be 
drawn  at  once,  since  the  corresponding  ray  is  known  ;  and  the 
other  is  known  after  the  remaining  strings  have  been  drawn, 
since  it  must  close  the  polygon.  The  construction  should  be 
carefully  followed  through  by  the  student. 

The  funicular  polygons  for  the  two  directions  of  the  wind  are 
distinguished  by  the  use  of  O  and  Or  to  designate  the  two  poles. 


70  GRAPHIC   STATICS. 

In  Fig.  33  (A),  ABCDEFGHIA  is  the  force  polygon  for  the 
case  when  the  wind  is  from  the  right.  Notice  that  the  points 
A,  B,  C,  D,  coincide.  This  means  that  the  loads  AB,  BC,  CD, 
are  each  zero. 


(*) 


In  diagram  (B),  ABCDEFGHTA  is  the  force  polygon  for 
the  case  when  the  wind  is  from  the  left.  The  points  £,  F,  G,  //, 
coincide,  because  the  loads  EF,  FG,  GH,  are  each  zero. 

93.  Stress  Diagrams  for  Wind  Pressure.  —  When  the  loads 
and  reactions  due  to  wind  pressure  are  known,  the  internal 
stresses  can  be  found  by  drawing  a  stress  diagram,  just  as  in 
the  case  of  vertical  loads.  The  construction  involves  no  new 
principle,  and  will  be  readily  made  by  the  student.  In  Figs. 
33  (A)  and  33  (B)  are  shown  the  diagrams  for  the  two  directions 
of  the  wind. 


MAXIMUM    STRESSES.  ji 

The  stresses  in  all  members  of  the  truss  must  be  determined 
for  each  direction  of  the  wind.  If  the  truss  is  symmetrical 
with  respect  to  a  vertical  line,  as  is  usually  the  case,  it  may  be 
that  the  same  stress  diagram  will  apply  for  both  directions  of 
wind.  This  will  be  so  if  the  reactions  are  assumed  to  act  as 
in  cases  (i)  and  (2)  of  the  preceding  article.  In  the  case  repre- 
sented in  Fig.  33,  however,  the  hinging  of  one  end  of  the  truss 
destroys  the  symmetry  of  the  two  stress  diagrams,  and  both 
must  be  drawn  in  full. 

§  4.    Maximum   Stresses, 

94.  General  Principles.  —  For  the  purpose  of  designing  any 
truss  member,  it  is  necessary  to  know  the  greatest  stresses  to 
which  it  will  be  subjected  under  any  possible  combination  of 
loads. 

Stresses  are  combined  in  accordance  with  the  principle  stated 
in  Art.  78,  that  the  resultant  stress  in  a  truss  member  due  to 
the  combined  action  of  any  loads  is  equal  to  the  algebraic  sum 
of  the  stresses  due  to  their  separate  action. 

The  method  will  be  illustrated  by  the  solution  of  an  example 
with  numerical  data. 

95.  Problem  —  Numerical  Data.  —  Let  it  be  required  to  design 
a  wrought  iron  truss  of  40  ft.  span,  of  the  form  shown  in   Fig. 
34  (PI.  I).      Let  12  ft.  be  the  distance  apart  of  trusses,  and  let 
the  loads  be  as  follows  : 

Weight  of  truss,  to  be  assumed  in  accordance  with  the 
formula  of  Art.  82  :  W=%al(i+^l).  This  gives  I¥=i8oo\bs. 
Assuming  this  to  be  divided  equally  among  the  upper  panels, 
and  that  the  load  for  each  panel  is  borne  equally  by  the  two 
adjacent  joints,  the  load  at  each  of  the  joints  be,  cd,  de  is  450 
Ibs.  The  loads  at  the  end  joints  may  be  neglected,  being  borne 
directly  at  the  supports. 

Weight  of  roof . — This  depends  upon  the  materials  Used  and 
the  method  of  construction,  but  will  be  taken  as  6  Ibs.  per  sq.  ft. 


72  GRAPHIC    STATICS. 

of  roof  area,  giving  900  Ibs.  as  the  load  at  each  joint.  This, 
also,  is  a  permanent  load.  Total  permanent  load  per  joint, 
1350  Ibs. 

Weight  of  snozv.  —  Taking  this  as  15  Ibs.  per  horizontal 
square  foot,  we  find  1800  Ibs.  as  the  load  at  each  joint. 

Wind  pressure.  —  This  is  computed  from  the  formula 

A=_2sin«_  (Art.  85.) 

I  +sm2  a 

For  this  case  we  put  sin  a  =  -J|=|- ;  /„  — 40  Ibs.  per  sq.  ft.; 
whence  /a  =  35  Ibs.  per  sq.  ft.  (about).  This  gives  upon  each 
panel  of  the  roof  5250  Ibs.  Then  with  the  wind  from  either 
side,  the  wind  loads  on  that  side  would  be  2625  Ibs.,  5250  Ibs., 
2625  Ibs.  respectively. 

96.  Stress  Diagrams.  —  We  are  now  ready  to  construct  the 
stress  diagrams. 

The  truss  is  shown  (PL  I)  in-  Fig.  34  (A).  Fig.  34  (B)  is  the 
stress  diagram  for  permanent  loads.  No  diagram  for  snow 
loads  is  needed,  since  it  would  be  exactly  similar  to  that  for 
permanent  loads.  The  snow  load  at  any  joint  being  four-thirds 
as  great  as  the  permanent  load,  the  stress  in  any  member  due 
to  snow  is  four-thirds  that  due  to  permanent  loads. 

Fig.  34  (C)  shows  the  stress  diagram  for  the  case  of  wind 
blowing  from  the  left.  The  reactions  are  assumed  to  act  in 
lines  parallel  to  the  loads  —  that  is,  normal  to  the  roof.  With 
this  assumption,  no  separate  diagram  is  needed  for  the  case  of 
wind  from  the  right,  since  such  a  diagram  would  be  exactly 
symmetrical  to  Fig.  34  (C).  For  example,  the  stress  in  the 
member  £•//  due  to  the  wind  blowing  from  the  right  is  given  by 
the  line  GM  in  Fig.  34  (C). 

97.  Combination  of  Stresses.  — After  the  stress  diagrams  are 
completed  for  the  various  kinds  of  loads,  the  stresses  should 
be  scaled  from  the  diagrams  and  entered  with  proper  sign  in  a 
table,  as  follows  : 


l'$«:<L'$!jL 

. 

MAXIMUA 

STRESSES. 

Member. 

Permanent 
Load. 

Snow.     Wind  R. 

Wind  L. 

Max. 

bh 
ci 

-  1525 
-  1255 

—  2030 
-  1670 

-  6270 
—  6270 

-  7925 
-  7925 

—  11480 
—  10850 

hi 

-  360. 

-  480 

o 

-5250 

-  6090 

ik 

+  595 

+  790 

+  550 

-f  6650 

-  8035 

& 

+  1245 

+  1660 

+  2600 

+  8800 

+  H705 

7 

+  730 

+  975 

+  2300 

+  2300 

+  4005 

gift 

+  1245 

-f  1660 

+  8800 

+  2600 

+  II705 

Ik 

+  595 

-f  790 

+  6650 

+  55° 

-  8035 

ml 

360 

-  480 

-5250 

o 

-  6090 

dl 

-  1255 

-  1670 

-7925 

—  6270 

—  10850 

em 

-  ^25 

—  2030 

-  7925 

—  6270 

-  11480 

73 


By  combining  the  results,  the  maximum  stress  in  each  mem- 
ber for  any  possible  condition  of  loading  can  be  determined. 
The  possible  combinations  of  loading  are  the  following  :  Perma- 
nent load  alone ;  permanent  and  snow  loads ;  permanent  load, 
and  wind  from  either  direction  ;  permanent  and  snow  loads,  and 
wind  from  either  direction.  The  student  will  readily  under- 
stand the  method  of  combining  the  separate  results. 

The  problem  here  solved  relates  to  a  very  simple  form  of 
truss.  With  some  forms  there  may  occur  a  reversal  of  stress 
in  certain  members,  under  different  conditions  of  loading. 

It  is  to  be  noticed  that  in  the  table  the  word  maximum  is 
used  in  its  numerical  sense,  and  has  no  reference  to  the  algebraic 
sign  of  the  stress. 

98.  Examples.  —  In  Fig.  35  (A)  to  (F),  are  shown  several 
forms  of  truss  for  which  the  student  may  draw  stress  diagrams, 
assuming  loads  in  accordance  with  the  data  given  in  Arts.  82 
to  85.  In  determining  reactions  due  to  wind  pressure,  the 


74 


GRAPHIC    STATICS. 


three  assumptions  mentioned  in  Art.  92  should  all  be  used  in 
different  cases,  that  the  student  may  become  familiar  with  the 
principle  of  each. 


(A) 


jrig.35 

§   5.    Cases  Apparently  Indeterminate. 

99.  Failure  of  Usual  Method.  —  In  attempting  to  construct 
the  stress  diagram  by  drawing  the  force  polygon  for  each  joint 
in  succession,  as  in  the  cases  thus  far  treated,  a  difficulty  is 
met  in  certain  forms  of  truss.  It  may  happen  that  after  pro- 
ceeding to  a  certain  point  it  is  impossible  to  select  a  joint  for 
which  the  force  polygon  can  be  completely  drawn,  the  number 
of  unknown  forces  for  every  joint  being  greater  than  two. 

Thus,  in  the  truss  shown  in  Fig.  36,  if  the  stress  diagram  is 
started  in  the  usual  way,  beginning  at  the  left  support,  the  force 
polygons  for  three  joints  may  be  constructed  without  difficulty, 
thus  determining  the  stresses  in  bl,  Ik,  Im,  cm,  mn,  nk.  But 
the  force  polygon  for  cdqpnmc  cannot  be  constructed,  since 
three  forces  are  unknown,  —  namely,  those  in  dq,  qp,  pn.  And 
at  the  joint  knpsk,  the  stresses  in  ;//,  ps,  and  sk  are  unknown. 
The  problem,  therefore,  seems  at  this  point  to  become  indeter- 


CASES  APPARENTLY  INDETERMINATE.         75 

minate,  since  either  of  the  two  polygons  can  be  completed  in 
any  number  of  ways,  so  far  as  the  known  forces  determine.  It 
can  be  shown,  however,  that  this  ambiguity  is  only  apparent. 
This  may  be  proved  as  follows  : 

Consider  the  portion  of  the  truss  to  the  left  of  the  broken 
line  M'N'.  It  is  in  equilibrium  under  the  action  of  eight 
forces  ;  five  of  these  (four  loads  and  a  reaction)  are  known  ; 
the  remaining  three  are  the  forces  in  cr,  rs,  and  sk.  Now,  the 
problem  of  determining  these  three  unknown  forces  is  the  same 
as  that  treated  in  Art.  40.  It  was  there  found  to  be  a  deter- 
minate problem,  unless  the  lines  of  action  of  the  three  unknown 
forces  intersect  in  a  point  or  are  parallel. 

That  the  problem  is  determinate  may  be  seen  also  from  the 
principle  of  moments  (Art.  51).  The  eight  forces  mentioned 
being  in  equilibrium,  the  sum  of  their  moments  is  zero  for  any 
origin  in  their  plane.  Let  the  origin  be  taken  at  the  point  of 
intersection  of  the  lines  of  action  of  two  of  the  unknown  forces, 
as  cr,  rs.  Then  from  the  principle  of  moments  we  have  (since 
the  moments  of  the  two  forces  named  are  zero) :  Algebraic  sum 
of  moments  of  loads  and  reaction  to  left  of  section  -f  moment 
of  SK=o.  The  only  unknown  quantity  in  this  equation  is  the 
magnitude  of  SK,  which  may,  therefore,  be  determined.  The 
other  unknown  forces  may  be  found  in  a  similar  manner, 
the  origin  of  moments  being  in  each  case  chosen  so  as  to 
eliminate  two  of  the  three  unknown  forces. 

The  whole  problem  of  drawing  the  stress  diagram  is  now 
seen  to  be  determinate.  For,  as  soon  as  the  stress  in  sk  is 
known,  the  force  polygon  for  the  joint  knpsk  contains  but  two 
unknown  sides,  and  can  be  drawn  at  once.  No  further  diffi- 
culty will  be  met. 

100.    Solution    of    Case   of    Failure  —  First    Method. — The 

reasoning  of  the  preceding  article  suggests  two  methods  of 
treating  the  so-called  ambiguous  case.  These  will  now  be 
described. 


y6  GRAPHIC   STATICS. 

The  first  method  is  to  apply  the  construction  of  Art.  40,  as 
follows  :  Referring  to  Fig.  36,  consider  the  equilibrium  of  the 
portion  of  the  truss  to  the  left  of  the  line  M'N'.  The  system 
of  forces  consists  of  those  whose  lines  of  action  are  ka,  ab,  be, 
cd,  de,  er,  rs,  sk.  Let  them  be  taken  in  the  order  named,  and 
draw  the  force  polygon  for  the  known  forces.  (The  reaction 
ka  is  supposed  to  be  already  determined.)  The  known  part  of 
the  force  polygon  is  KABCDE\  the  unknown  part  is  to  be 
marked  ERSK.  Choose  a  pole  O  and  draw  rays  to  K,  A,  B, 
C,  D,  and  E\  then  draw  the  corresponding  strings  of  the  funic- 
ular polygon.  Remembering  the  method  of  Art.  40,  we  draw 
first  oe,  making  it  pass  through  the  intersection  of  cr  and  rs 


(the  reason  for  this  being  that  it  makes  the  string-  os  also  pass 
through  that  point]  ;  then  draw  in  succession  od,  ocy  ob,  oa,  ok. 
The  string  oh  intersects  sk  in  a  point  through  which  os  must  be 
drawn.  Hence  os  must  join  that  point  with  the  starting  point 
of  the  polygon.  This  completes  the  funicular  polygon,  except 
the  string  or,  which  must  pass  through  the  intersection  of  er 


CASKS  APPARENTLY  INDETERMINATE.        77 

and  rs  in  some  direction  not  yet  known.  This  string  is  not 
necessary  to  the  solution  of  the  problem. 

Since  os  is  now  known,  OS  may  be  drawn  parallel  to  it  from 
O ;  and  by  drawing  a  line  from  K  parallel  to  the  known  direc- 
tion of  KS,  the  point  5  is  determined,  and  KS  becomes 
known. 

To  find  ER  and  RS  it  is  only  necessary  to  draw  from  5  a 
line  parallel  to  rs,  and  from  E  a  line  parallel  to  er ;  their  inter- 
section gives  the  point  R,  and  the  force  polygon  for  the  system 
of  forces  considered  is  complete.  The  stresses  in  er,  rs,  and  sk 
being  now  known,  the  stress  diagram  may  be  completely  drawn 
by  the  usual  method. 

It  is  interesting  to  notice  that  the  method  just  described 
determines  the  lines  ER,  RS,  SK,  in  their  proper  position  in 
the  complete  stress  diagram.  The  determination  of  ER  and 
RS  by  this  method  is  not  necessary,  since  the  usual  method  of 
drawing  the  stress-diagram  can  be  carried  out  as  soon  as  SK 
is  known.  But  the  independent  determination  of  ER  and  RS 
by  the  two  methods  serves  as  a  check  on  the  accuracy  of  the 
construction. 

The  funicular  polygon  employed  in  the  above  construction, 
so  far  as  it  belongs  to  the  external  forces,  may  coincide  with 
the  corresponding  part  of  the  funicular  polygon  used  in  deter- 
mining the  reactions.  If  this  is  desired,  two  points  must  be 
observed:  (i)  the  string  oe  must  be  the  first  drawn,  and  must 
pass  through  the  intersection  of  er  and  rs,  and  (2)  the  pole 
must  be  so  chosen  that  ok  will  not  be  nearly  parallel  to  ks. 

It  should  also  be  noticed  that  the  construction  of  the  funic- 
ular polygon  might  begin  with  the  string  ok,  which  should  then 
be  made  to  pass  through  the  intersection  of  rs  and  sk.  The 
student  will  be  able  to  carry  out  this  construction  without 
difficulty. 

10 1.    Solution  of  Case  of  Failure  —  Second  Method.  —  It  will 

now   be    shown    how  the   apparently  ambiguous    case   can    be 


78  GRAPHIC   STATICS. 

treated  graphically  by  the  principle  of  moments.  Referring 
again  to  Fig.  36,  consider  the  portion  of  the  truss  to  the  left 
of  section  M'JV'.  It  is  acted  upon  by  eight  forces,  of  which 
five  (the  loads  and  the  reaction)  are  known,  and  three  (whose 
lines  of  action  are  er,  rs,  sk)  are  unknown.  The  algebraic  sum 
of  the  moments  of  all  these  forces  about  any  origin  must  be 
zero.  Let  the  origin  be  taken  at  the  point  of  intersection  of 
er  and  rs,  so  that  the  moments  of  the  forces  acting  in  these 
lines  are  both  zero ;  then  the  sum  of  the  moments  of  the  five 
known  forces,  plus  the  moment  of  the  force  acting  in  the  line 
sk,  must  equal  zero.  Now  the  sum  of  the  moments  of  the 
five  known  forces  may  be  found  by  the  method  of  Art.  56. 
Through  the  origin  of  moments  draw  a  line  parallel  to  the 
resultant  of  the  forces  named  (that  is,  a  vertical  line),  and  let 
i  equal  the  length  intercepted  on  it  by  the  strings  oe,  ok.  Then 
the  required  moment  is  —  iH,  where  H  is  the  pole  distance. 
(The  minus  sign  is  given  in  accordance  with  the  convention 
that  left-handed  rotation  shall  be  positive.)  Let  P  =  unknown 
force  in  line  sk,  and  h  its  moment-arm.  For  the  purpose  of 
computing  the  moment,  assume  the  stress  in  sk  to  be  a  tension  ; 
then  the  force  P  acts  toward  the  right  and  its  moment  is  posi- 
tive, the  value  being  +  Ph. 

Hence,  Ph-Hi=o\ 

or,  P=l-H. 

h 

From  this  equation  P  may  be  computed.  The  computation 
may  be  made  graphically  as  follows  :  Draw  (Fig.  36)  a  triangle 
WUV,  making  WU=H  (force  units)  and  WV=h  (linear 
units).  Lay  off  WY—i  (linear  units)  and  draw  YX  parallel 
to  VU.  Then  WX  (force  units)  represents  P.  This  is  readily 

seen,  since  from  the  two  similar  triangles  we  have  the  propor- 

P      i 

tion  —  =  -,  which  agrees  with  the  equation  above  deduced. 
H      Ji 

The  computation  is  simplified  if  the  pole  distance  H  is  taken 
equal  to  as  many   force   units    as  //  is  linear  units  ;    or  if   H 


CASES   APPARENTLY    INDETERMINATE.  79 

is    some    simple    multiple    of    h.     For,   suppose    H=nh\    then 

P'"S  A      m' 

The  stress  in  sk  is  found  to  be  a  tension,  since  P  is  positive. 

Whatever  the  nature  of  the  stress,  it  may  be  assumed  a  ten- 
sion in  writing  the  equation,  and  the  sign  of  the  value  found 
will  show  whether  the  assumption  coincides  with  the  fact. 

1 02.  Other  Methods  for  Case  of  Failure.  —  In  certain  cases 
the  method  of  treating  the  "  ambiguous  case  "  may  profitably 
be  varied. 

(1)  The    construction  of   Art.    100  may  be  modified  as  fol- 
lows :  Determine  the   resultant  of    the  known  external  forces 
acting  on  the  portion  of  the  truss  to  one  side  of  the  section 
M'N'.     This  resultant  is  in  equilibrium  with  the  three  unknown 
forces  acting  in  the  members  er,  rs,  sk.     Hence  these  forces  can 
be  determined  by  the  special  method  explained  in  Art.  42. 

The  resultant  of  the  five  known  forces  is  represented  in 
magnitude  and  direction  by  KE  in  the  force  polygon ;  and  its 
line  of  action  passes  through  the  intersection  of  the  strings  oe, 
ok  in  the  funicular  polygon.  Since  this  point  of  intersection  is 
likely  to  be  inaccessible,  the  construction  of  Art.  42  cannot  be 
conveniently  applied.  It  may  be  modified  by  using  instead  of 
KE  the  two  forces  KA  (the  reaction  in  the  line  ka)  and  AE 
(the  resultant  of  the  four  loads,  its  line  of  action  being  deter- 
mined by  the  intersection  of  the  strings  oa  and  oe).  First 
determine  forces  in  the  three  lines  er,  rs,  sk,  which  shall  be  in 
equilibrium  with  KA  ;  then  make  a  similar  construction  for 
AE,  and  combine  the  results. 

(2)  It  has  been  proposed  to  employ  the  following  reasoning : 
Remove  the  members/^  and  qr,  and  insert  another  represented 
by  the  broken  line  in  Fig.  36.     Evidently  this  does  not  change 
the    stress  in  the  member  sk,  since  the  forces  acting  on  the 
truss  to  the  left  of  the  section  M'N'  are  unchanged.     But  with 
this  change  the  difficulty  encountered  in  constructing  the  stress 
diagram  by  the  usual  method  is  avoided.     For  when   the  joint 


8o  GRAPHIC   STATICS. 

nmcdqpn  is  reached,  the  forces  acting  there  will  be  all  known, 
except  two.  Let  the  stress  diagram  be  drawn  in  the  usual  way 
until  the  stress  in  sk  is  known.  Then  restore  the  original 
bracing  and  repeat  the  construction,  using  the  value  just  deter- 
mined for  SK. 

This  method  is  convenient  whenever  it  is  applicable.  Cases 
may,  however,  arise,  in  which  it  will  fail.  For  instance,  if 
a  load  is  applied  at  the  joint  pqrsp,  the  members  pq  and  qr 

cannot    both    be    omitted,    and 
the  method  cannot  be  applied. 
Again,  it  would  be  inapplicable 
Fis' 37  in  case  of  a  truss  such  as  shown 

in  Fig.  37,  with  loads  at  all  upper  joints.     In  such  cases,  one  of 
the  methods  explained  in  the  preceding  articles  may  be  applied. 
Other  methods  might  be  mentioned,  but  the  foregoing  dis- 
cussion of  the  case  will  probably  be  found  sufficient. 

103.  Failing   Case   in   Other   Forms  of   Truss.  —  The  usual 
method   of  constructing  the   stress  diagram  may  fail  in  other 
forms  of  truss,  though   the  one  above  described  is  the  most 
common.     In  such  a  case,  the  problem  of  finding  the  stresses 
may  be  really  indeterminate,  or  only  apparently  so.     Whenever 
it  is  possible  to  divide  the  truss  into  two  parts  by  cutting  three 
members  which   are  not  parallel  and  do  not  intersect  in  one 
point,  the  stresses  in  the  three  members  cut  are  determinate  as 
soon  as  all  external   forces  are  known,  and  can  be  found  by 
methods  already  given.     If  more  than  three  members  are  cut, 
the  problem  of  finding  the  stresses  in  them  is  indeterminate, 
unless  all  but  three  of  these  stresses  are  known.     By  remem- 
bering these  principles,  the  determinateness  of  any  given  prob- 
lem may  readily  be  tested.     (See  Art.  70.) 

§  6.     Three-Hinged  Arch. 

104.  Arched  Truss  Defined.  —  If  a  truss  is  so  supported  that 
when   sustaining  vertical  loads   the  reactions  at  the  supports 


THREE-HINGED   ARCH. 


8l 


have  horizontal  components  directed  toward  the  centre  of  the 
span,  the  truss  becomes  an  arcJi.  The  only  kind  of  arch  we 
shall  here  consider  is  that  consisting  of  two  partial  trusses 
hinged  together  at  the  crown,  and  each  hinged  at  the  point  of 
support. 

Such  a  truss  is  shown  in  Fig.  38,  in  which  the  two  partial 
trusses  are  hinged  to  the  abutments  at  P  and  Q,  and  connected 
by  a  hinge  at  the  point  R.  Since  a  hinge  at  the  support  allows 
the  reaction  of  the  supporting  body  upon  the  truss  to  take  any 
direction  in  the  plane  of  the  truss,  the  directions  of  the  reactions 
at  P  and  Q  are  unknown,  as  is  also  that  of  the  force  exerted  by 
either  partial  truss  upon  the  other  at  the  point  R.  The  problem 
of  determining  the  reactions  may,  therefore,  at  first  sight  seem 
indeterminate.  It  will  tye  shown  in  the  next  article  that  it  is 
in  reality  determinate,  and  that  the  only  principles  needed  in 
the  solution  are  such  as  have  been  already  often  applied  in  the 
preceding  chapters.  The  three-hinged  arch  is  indeed  a  "  simple  " 
structure  (Art.  75),  since  the  theory  of  elasticity  is  not  needed 
in  the  determination  of  the  reactions. 

105.  Reactions  Due  to  a  Single  Load. — The  method  of  find- 
ing the  reactions  is  most  clearly  understood  by  considering  the 


CF) 


effect  of  a  single  load  on  either  partial  truss.  Let  a  load  be 
applied  at  S  (Fig.  38)  and  let  all  other  loads  acting  on  either 
portion  of  the  truss  (including  the  weight  of  the  structure)  be 


82  GRAPHIC   STATICS. 

neglected.  Call  the  two  partial  trusses  X  and  F,  and  consider 
the  part  F  The  Only  forces  acting  upon  it  are  the  reaction 
exerted  by  the  abutment  at  Q  and  the  force  exerted  at  R  by 
the  truss  X.  These  two  forces,  being  in  equilibrium,  must 
have  the  same  line  of  action,  which  is,  therefore,  the  line  QR. 
Consider  now  the  body  X.  The  external  forces  acting  upon  it 
are  the  load  at  S,  the  reaction  of  the  abutment  at  P,  and  the 
force  exerted  by  Fat  the  point  R.  But  this  last  force  is  equal 
and  opposite  to  the  force  exerted  by  X  upon  F,  and  its  line  of 
action  is  therefore  QR.  The  three  forces  acting  upon  the 
body  X  being  in  equilibrium,  their  lines  of  action  must  meet  in 
a  point ;  which  point  is  found  by  prolonging  QR  to  meet  the 
line  of  action  of  the  applied  load.  Let  T  be  this  point,  then 
PT  is  the  line  of  action  of  the  reaction  at  P.  The  reactions 
can  now  be  determined  by  drawing  the  triangle  of  forces.  This 
triangle  is  ABC  in  the  figure,  AB  representing  the  load  at  5,  BC 
the  reaction  in  the  line  QR  (also  marked  be),  and  CA  the  reac- 
tion in  the  line  PT  (marked  also  ca).  Evidently  ABC  may  be 
regarded  as  the  polygon  of  external  forces,  either  for  the  partial 
truss  X,  or  for  the  whole  structure  composed  of  X  and  F;  and 
BC  represents  either  the  force  exerted  by  F  upon  X  at  R,  or 
the  force  exerted  upon  F  by  the  supporting  body  at  Q. 

If,  now,  the  structure  be  loaded  at  other  points,  the  reactions 
due  to  each  load  may  be  found  separately ;  the  resultant  of 
all  such  separate  reactions  at  either  support  will  be  the  true 
reaction  at  that  support  when  all  the  loads  act  together.  A 
convenient  method  of  applying  these  principles  will  be  given  in 
the  next  article. 

1 06.   Reactions  and  Stresses  Due  to  Any  Vertical  Loads.  —  In 

Fig.  39  is  represented  a  truss  consisting  of  two  parts  supported 
by  hinges  at  P'  and  Q  and  hinged  together  at  R' .  Consider 
all  vertical  loads  to  be  applied  at  the  upper  joints,  their  lines  of 
action  being  marked  in  the  usual  way.  We  shall  first  explain 
the  construction  for  finding  the  reactions  at  the  supports  ;  after 


THREE-HINGED   ARCH.  83 

these  are  determined,  the  drawing  of  the  stress  diagram  will 
present  no  difficulty. 

Since  we  shall  sometimes  deal  with  one  of  the  partial  trusses, 
and  sometimes  with  the  two  considered  as  a  single  body,  it  will 
be  well  at  the  outset  to  specify  the  external  forces  acting  on 


each  of  these  bodies.     X 


(B) 


(i)  For  the  partial  truss  at  the  left  we  have  five  applied  loads, 
the  reaction  at  R'  (exerted  by  the  other  truss),  and  the  reaction 
at  P'.  The  force  polygon  for  these  forces  will  be  marked  as 
follows:  ABCDEFLA.  (The  meaning  of  the  letters  will  be 
understood  before  the  force  polygon  is  actually  drawn,  by 
reference  to  the  corresponding  letters  in  the  space  diagram.) 


84  GRAPHIC    STATICS. 

(2)  For  the  right-hand    partial  truss  the  external  forces  are 
five   loads   and   two   reactions,   and  the  force   polygon   will   be 
marked  thus  :  FGHIJKLF.      (Notice  that  FL  and  ZFare  equal 
and  opposite  forces,  being  the  "action  and  reaction"  between 
the  two  trusses  at  7?'.) 

(3)  For  the  combined  structure  the  external  forces  are  the 
ten  loads  and  the  reactions  at  Pr  and   Q'.     (The  action   and 
reaction  at  R  become  now  internal  forces.)     The  force  polygon 
will  be  marked  thus  :  ABCDEFGHIJKLA, 

Begin  the  construction  by  drawing  the  force  polygon  for  the 
ten  loads  on  the  whole  structure,  lettering  it  as  just  indicated. 
Choose  a  pole  O,  draw  the  rays,  and  then  the  funicular  polygon 
as  far  as  possible.  Now  consider  the  right  partial  truss  as 
unloaded.  The  resultant  of  the  remaining  five  loads  is  repre- 
sented in  magnitude  and  direction  by  AF\  its  line  of  action 
must  pass  through  the  intersection  of  oa  and  of,  and  is  therefore 
the  line  marked  af.  Now,  reasoning  as  in  the  preceding  article, 
we  see  that  the  reaction  at  Q'  must  act  in  the  line  Q'R'.  Let 
Q'R1  intersect  af  in  T' ,  then  P1  T  is  the  line  of  action  of 
the  reaction  at  P'.  Hence  the  complete  force  polygon  for  the 
whole  truss,  when  the  right  half  is  unloaded,  may  be  found  by 
drawing  from  Fa  line  parallel  to  Q'R',  and  from  A  a  line  parallel 
to  P' T',  prolonging  them  till  they  intersect  at  L' .  The  reaction 
at  P1  is  L'A,  and  that  at  Q'  is  FL1.  (The  line  of  action  of  the 
latter  is  marked//7.) 

Next,  consider  the  left  half  to  be  without  loads,  the  other 
half  being  loaded.  The  resultant  of  the  five  loads  now  acting  is 
FK,  its  line  of  action  fk  being  drawn  through  the  point  of 
intersection  of  of  and  ok.  The  reactions  at  P'  and  Q'  for  the 
present  case  .have  lines  of  action  P' R'  and  Q'T",  found  just 
as  in  the  first  case  of  loading.  These  reactions  are  therefore 
determined  in  magnitude  and  direction  by  drawing  from  A'  a 
line  parallel  to  Q'T"  and  from  F  a  line  parallel  to  P'R',  pro- 
longing them  till  they  intersect  at  L".  The  complete  force 
polygon  for  this  case  of  loading  is  therefore  FGHIJKL"F. 


THREE-HINGED    ARCH.  85 

Consider  now  that  both  parts  of  the  truss  are  loaded.  The 
reaction  at  P'  is  the  resultant  of  the  two  partial  reactions  L'A, 
L"F,  and  the  reaction  at  Q'  is  the  resultant  of  the  two  partial 
reactions  FLr,  KL" .  From  Ln  and  L'  draw  lines  parallel 
respectively  to  FL'  and  FL",  intersecting  in  L.  Then  L" L  is 
equal  and  parallel  to  FL' ,  and  LL1  is  equal  and  parallel  to  L"F. 
Hence  KL  and  LA  represent  the  resultant  reactions  at  Q'  and 
P'  respectively.  This  completes  the  polygon  of  external  forces 
for  the  whole  truss,  as  well  as  that  for  each  partial  truss. 

The  stress  diagram  can  now  be  drawn  in  the  usual  way, 
beginning  at  the  point  P'.  The  diagram  for  one  partial  truss 
is  shown  in  Fig.  39  (B). 

If  loads  are  applied  at  the  lower  joints  of  the  trusses,  the 
reactions  due  to  these  may  be  found  in  the  same  manner  as  for 
the  upper  loads.  But  before  beginning  the  determination  of 
the  stresses,  the  polygon  must  be  drawn  for  the  external  forces 
taken  in  order  around  the  truss.  (See  Art.  90.) 

107.  Case  of  Symmetrical  Loading.  —  If  the  two  half  trusses 
are  exactly  similar  and  symmetrically  loaded,  the  determination 
of  reactions  and  stresses  is  much  simplified. 

(1)  As  regards  the  reactions,  symmetry  shows  that  the  forces 
exerted  by  the  two  trusses  upon  each  other  at  R'  are  horizontal. 
Hence,  referring  to  Fig.  39,  and  considering  either  half-truss, 
as  that  to  the  left,  the  line  of  action  of  the  reaction  at  P'  may 
be  found  by  drawing  a  horizontal  line  through  R1  and  prolong- 
ing it  to  intersect  of,  the  line  of  action  of  the  resultant  of  all 
loads  on  the  left  truss  ;  the  line  joining  this  point  of  intersection 
with  P'  is  the  required  line  of  action  of  the  reaction  at  the  left 
abutment.     The  two  reactions  are  now  determined  by  drawing 
from  Fa.  horizontal  line  and  from  A  a  line  parallel  to  the  line 
just  determined,  and  prolonging  them  till  they  intersect. 

(2)  As  to  the  stresses,  only  one  partial  truss  need  be  consid- 
ered, since  the  stress    diagrams  for  the  two  portions  will  be 
symmetrical  figures. 


86 


GRAPHIC   STATICS. 


These  principles  might  have  been  employed  in  the  case  dis- 
cussed in  the  preceding  article ;  but  the  method  there  used  is 
applicable  to  cases  in  which  neither  the  trusses  nor  the  loads 
are  symmetrical. 

108.  Wind  Pressure  Diagram.  —  The  diagrams  for  wind 
pressure  will  present  no  difficulty.  The  determination  of  the 
reactions  will,  indeed,  be  simpler  than  in  the  preceding  case, 
since  only  one  partial  truss  will  be  loaded  at  any  one  time,  and 
the  line  of  action  of  one  reaction  is  therefore  known  at  the 
outset.  The  construction  is  shown  in  Fig.  40. 


f\k 


Fig.  40. 


In  computing  wind  pressure  loads,  it  will  be  assumed  that 
each  joint  sustains  half  the  pressure  coming  upon  each  of  the 
two  adjacent  panels.  It  will  be  sufficiently  correct  in  comput- 
ing the  pressure  at  any  joint,  to  assume  the  slope  of  the  roof 


THREE-HINGED   ARCH.  87 

as  that  of  the  tangent  to  the  roof  curve  at  the  joint  in  ques- 
tion ;  and  the  direction  of  the  wind  load  may  be  taken  as  that 
of  the  normal  to  the  roof  curve  at  the  joint.  The  force  poly- 
gon for  the  wind  loads  is  therefore  not  a  straight  line  when  the 
roof  is  curved.  In  Fig.  40,  this  polygon  is  FGHIJK.  The 
reactions  are  to  be  marked  KL,  LF.  (Since  there  are  no  loads 
on  the  other  half-truss,  the  points  ABCDEF  will  coincide.) 
Choosing  a  pole  O,  draw  the  funicular  polygon  as  shown,  and 
determine  the  line  of  action  of  the  resultant  of  all  the  wind 
loads.  This  line  of  action />£  is  drawn  parallel  to  FK,  through 
the  point  of  intersection  of  the  strings  of  and  ok.  Prolong /£ 
to  intersect  P'R'  produced  at  T" ;  then  Q'T"  is  the  line  of 
action  of  the  reaction  at  Q'.  The  force  polygon  may  now  be 
completed  by  drawing  KL  parallel  to  Q'T'  and  LF  parallel  to 
P'R1.  The  reactions  KL  and  LF  being  thus  determined,  the 
stress  diagram  can  be  drawn  without  difficulty. 

The  stress  diagram  is  drawn  for  both  partial  trusses,  with  the 
result  shown  in  the  figure.  If  the  two  trusses  are  symmetrical, 
the  diagram  for  the  other  direction  of  the  wind  need  not  be 
drawn ;  the  stresses  for  this  case  being  found  from  the  diagram 
already  drawn.  If,  however,  the  partial  trusses  are  dissimilar,  a 
second  wind-diagram  must  be  drawn. 

It  is  not  necessary  to  draw  separate  space  diagrams  for  verti- 
cal loads  and  wind-forces.  The  constructions  of  Figs.  39  and 
40  have  been  here  kept  wholly  separate,  in  order  that  the 
explanation  may  be  more  easily  followed. 

109.  Check  by  Method  of  Sections.  —  In  case  of  a  truss  of 
long  span,  especially  when  the  members  have  many  different 
directions  and  are  short  compared  with  the  whole  length  of  the 
span,  the  small  errors  made  in  drawing  the  stress  diagram  are 
likely  to  accumulate  so  much  as  to  vitiate  the  results.  Thus, 
in  Fig.  39,  if,  in  drawing  the  stress  diagram,  we  begin  at  P'  and 
pass  from  joint  to  joint,  there  is  no  check  upon  the  correctness 
of  the  work  until  the  point  R'  is  reached.  At  that  point  we 


88  GRAPHIC    STATICS. 

have  a  second  determination  of  the  reaction  exerted  by  each 
half-truss  upon  the  other ;  and  it  is  quite  likely  that  the  two 
values  found  will  not  agree. 

By  a  method  like  that  employed  in  Art.  100  for  the  "  inde- 
terminate" form  of  truss,  we  may  avoid  the  necessity  of  mak- 
ing so  long  a  construction  before  checking  the  results.  In 
Fig.  39  take  a  section  cutting  the  three  members  cq,  qr,  rl, 
and  apply  the  principles  of  equilibrium  to  the  body  at  the 
left  of  the  section.  The  forces  acting  on  this  body  are  LA, 
AB,  BC,  CQ,  QR,  RL.  Choose  a  pole  O'  and  draw  the  funic- 
ular polygon  for  this  system,  making  the  string  o'c  pass  through 
the  point  of  intersection  of  cq  and  qr.  Draw  successively  the 
strings  o'c,  o'b,  o'a,  o'/,  prolonging  the  last  to  intersect  /;-. 
Through  the  point  thus  determined,  draw  e'r,  which  must  also 
pass  through  the  point  of  intersection  of  cq  and  qr.  As  soon  as 
o'r  is  known,  the  corresponding  ray  O'R  can  be  drawn  in  the 
force  diagram,  and  then  by  drawing  from  L  a  line  parallel  to 
Ir,  the  point  R  is  determined.  We  may  now  close  the  force 
polygon,  since  the  directions  of  the  two  remaining  forces  (cq 
and  qr)  are  known. 

The  stresses  in  the  three  members  cut  being  now  known,  we 
have  a  check  on  the  correctness  of  the  construction  of  the 
stress  diagram,  as  soon  as  these  members  are  reached  in  the 
process. 

This  method  will  be  found  of  great  use,  not  only  for  this 
form  of  truss,  but  for  any  truss  containing  many  members. 

§  7.    Connterbracing. 

no.  Reversal  of  Stress.  — If  the  loads  supported  by  a  truss 
are  fixed  in  position  and  unchanging  in  amount,  the  stress  in 
any  member  remains  constant  in  magnitude  and  kind.  But  in 
most  cases  such  are  not  the  conditions,  and  it  may  happen  that 
under  different  combinations  of  loading,  the  stress  in  a  certain 
member  is  sometimes  tension  and  sometimes  compression.  It 


COUNTERBRACING.  89 

is  often  thought  desirable  to  prevent  such  changes  of  stress, 
the  design  of  the  members  and  their  connections  being  thereby 
simplified.  To  accomplish  this  is  the  object  of  counterb racing. 

iii.  Counter  bracing.  —  Consider  a  truss  such  as  the  one 
shown  in  Fig.  41,  subjected  to  vertical  loads  and  to  wind  pres- 
sure from  either  side.  A  diagonal  member  such  as  xy  may 
sustain  tension  under  vertical  loads  alone,  or  with  the  wind 
from  the  left ;  while  with  the  wind  from  the  right  it  may  sustain 
compression. 

Now  suppose  xy  removed,  and  a  member  represented  by  the 
broken  line  xy'  introduced.  It  may  easily  be  shown  that  any 


'M 


system  of  loading  which  would  cause  compression  in  xy  will 
cause  tension  in  this  new  member ;  and  vice  versa.  For,  divide 
the  truss  by  a  section  MN  cutting  xy  and  two  chord  mem- 
bers as  shown,  and  let  L  be  the  point  of  intersection  of  the 
two  chord  members  produced.  The  kind  of  stress  in  xy  or 
the  other  member  (whichever  is  assumed  to  be  present)  may 
be  determined  by  considering  the  system  of  forces  acting  on 
one  portion  of  the  truss,  as  that  to  the  left  of  the  section  MN. 
Let  the  principle  of  moments  be  applied  to  this  system,  the 
origin  being  taken  at  L.  If  the  external  forces  acting  on  the 
portion  of  the  truss  considered  be  such  as  to  tend  to  cause 
right-handed  rotation  about  L,  the  stress  in  xy  must  be  com- 
pression in  order  to  resist  this  tendency  ;  while,  if  xy  be  replaced 
by  xy\  a  tension  must  exist  in  that  member  to  resist  right- 
handed  rotation  about  L.  Similarly,  a  tendency  of  the  external 
forces  to  produce  left-handed  rotation  about  L  would  be  resisted 
by  a  tension  in  xy ;  or  by  a  compression  in  the  member  xy' . 
If  the  two  members  act  at  the  same  time,  the  stresses  in 


90  GRAPHIC   STATICS. 

them  will  be  indeterminate.  These  stresses  may,  however,  be 
made  determinate  by  the  following  device  : 

Let  the  member  xy  be  so  constructed  that  it  cannot  sustain 
compression.  Then,  whenever  the  external  forces  are  such  as 
to  tend  to  throw  compression  upon  it,  it  ceases  to  act  as  a 
truss-member,  and  the  member  xy1  receives  a  tension  which  is 
determinate. 

If  the  member  xy1  be  constructed  in  the  same  way,  any 
tendency  to  throw  compression  upon  it  causes  it  immediately 
to  cease  to  act,  and  puts  upon  the  member  xy  a  tension  instead. 

A  member  such  as  xy' ,  constructed  in  the  manner  mentioned, 
is  called  a  counterbrace. 

112.   Determination   of  Stresses  with  Counterbracing.  —  The 

use  of  counterbracing  adds  somewhat  to  the  labor  of  deter- 
mining the  maximum  stresses,  since  the  members  actually  under 
stress  are  not  always  the  same.  The  method  of  treating  such 
cases  will  be  illustrated  in  the  next  article  by  the  solution  of  an 
example  ;  but  first  the  main  steps  in  the  process  may  be  outlined 
as  follows  : 

(a)  Construct  separate  stress  diagrams  for  vertical  loads  and 
for  wind  in  each  direction,  assuming  the  diagonals  in  all  panels 
to  slope  the  same  way. 

(b)  Determine  by  comparison  of  these  diagrams  in  which  of 
the  diagonal  members  the  resultant  stress  is  ever  liable  to  be  a 
compression.     Draw  in  counters  to  all  such  members. 

(c)  With   these    counterbraces    substituted    for   the    original 
members,  either  draw  new  stress  diagrams,  or  make  the  neces- 
sary additions  to  those  already  drawn.     If  the  latter  method  is 
adopted,  the  added  lines  should  be  inked  in  a  different  color 
from  that  employed  originally.     (In  some  cases,  this  construc- 
tion may  be  unnecessary  on  account  of  symmetry,  as  will  be 
illustrated  in  the  next  article.) 

(d)  Combine  the  separate  stresses  for  maxima  in  the  usual 
way. 


COUNTERBRACING.  91 

113.  Example.  —  In  Fig.  42  (PL  II)  are  given  stress  diagrams 
for  a  "  bow-string"  roof-truss  in  which  counterbracing  is  em- 
ployed. At  (A)  is  shown  the  truss  or  space  diagram.  The 
span  is  48  ft.  ;  rise  of  top  chord,  16  ft.  ;  rise  of  bottom  chord, 
8  ft.  The  chords  are  arcs  of  parabolas.  The  whole  truss  is 
divided  into  six  panels  by  equidistant  vertical  members.  The 
distance  apart  of  trusses  is  taken  as  12  ft. 

Assume  the  weight  of  the  truss  at  2400  Ibs.,  and  that  of  the 
roof  at  3600  Ibs. ;  this  makes  the  total  permanent  load  1000  Ibs. 
per  panel.  Take  800  Ibs.  as  the  load  at  each  upper  joint,  and 
200  Ibs.  as  the  load  at  each  lower  joint. 

The  snow  load,  computed  at  15  Ibs.  per  horizontal  square 
foot,  is  about  1440  Ibs.  per  panel. 

Wind  pressure  is  to  be  computed  from  the  formula  of  Art.  85. 

We  now  proceed  to  apply  the  method  outlined  in  the  preced- 
ing article. 

(a)  Assuming  one  set  of  diagonals  present,  we  construct  the 
stress  diagrams  for  the  various  kinds  of  loads. 

Diagram  for  permanent  loads.  —  This  is  shown  at  (B]  Fig. 
42,  which  needs  little  explanation.  It  will  be  noticed  that  the 
stress  in  every  diagonal  member  is  zero.  This  will  always  be 
the  case  if  the  chords  are  parabolic  and  the  vertical  loads  are 
equal  and  spaced  at  equal  horizontal  distances. 

Diagram  for  snow  loads.  —  Fig.  42  (C)  is  the  stress  diagram 
for  snow  loads.  In  this  case,  also,  the  stresses  in  the  diagonals 
are  all  zero. 

Wind  diagrams.  —  At  (D)  and  (E)  are  shown  the  diagrams 
for  the  two  directions  of  the  wind.  The  only  thing  needing 
special  mention  is  the  method  used  in  laying  off  the  force- 
polygon  for  the  wind  loads.  We  first  compute  the  normal  wind 
pressure  on  each  panel  by  the  formula  of  Art.  85.  We  thus 
find,  when  the  wind  is  from  the  right,  the  following  total  pres- 
sures, taking /„==  40  Ibs.  per  sq.  ft.  : 

On  panel  d,  1650  Ibs.    On  panel  e,  3870  Ibs.    On  panel/,  5480  Ibs. 


92 


GRAPHIC    STATICS. 


In  Fig.  42  (D)  these  are  laid  off  successively  in  their  proper 
directions  to  the  assumed  scale.  Thus  CD1,  D'E',  E'G  repre- 
sent respectively  1650  Ibs.  normal  to  dq,  3870  Ibs.  normal  to  cs, 
and  5480  Ibs.  normal  to  ft.  Now  each  of  these  loads  is  to  be 
equally  divided  between  the  two  adjacent  joints.  Bisect  CD'  at 
D,  D'E1  at  E,  and  E'G  at  F\  then  CD,  DE,  EF,  FG  represent 
the  loads  at  the  joints  cd,  de,  ef,  and  fg  respectively.  The 
"load  line"  is  therefore  CDEFG. 

The  reactions  are  assumed  to  be  parallel  to  the  resultant 
load.  With  this  explanation  the  figures  (D)  and  (E)  will  be 
readily  understood. 

(b)  Comparison  of  results.  —  The  stresses  due  to  permanent 
loads,  wind  right,  and  wind  left,  are  shown  in  tabular  form  for 
convenience  of  comparison. 


Member. 

Perm.  Load. 

Snow  Load. 

Wind  R. 

Wind  L. 

Max. 

aJ 

-  6660 

-9680 

-  4250 

-  14070 

—  16340 

bl 

-  535° 

-778o 

-  4880 

-  9650 

-  13*3° 

en 

-4550 

-6640 

-  6380 

-  62OO 

-  11190 

dq 

-4550 

-  6640 

-  95°° 

-  4250 

-  14050* 

es 

-535° 

-7780 

-  i5030 

-  345° 

-  20380* 

fl 

-  6660 

-  9680 

-  14070 

-  4300 

-  20730* 

A 

+  5080 

4-  7400 

+   780 

+  13500 

4-  12480 

khb 

+  4680 

+  6820 

-f   700 

+  12450 

4  11500 

m/i4 

-f  4470    +  6520 

+  !950 

+  7350 

+  10990 

P^ 

+  4470    +  6520 

+  4100 

+  4100 

+  10990* 

rh^ 

+  4680. 

4-  6820 

+  7680 

+  2050 

4-  12360* 

*, 

+  5080 

+  7400 

+  1350° 

+   800 

+  18580* 

jk 

+  1180 

+  1440 

+   150 

+  2650 

+  2620 

Im 

+  1180 

-f  1440 

-   280 

+  4100 

+  2620 

np 

+  1180 

+  1440 

95° 

+  3800 

+  2620* 

qr 

+  1180 

+  1440 

•  1650 

+  2525 

f  +  2620*  ) 
1  -   470*  J 

f  4-  2620*  | 

st 

+  1180    +  1440 

-  !35° 

+  1250 

1              T  *.,-,*   \ 

v.  —   170*  ' 

kl 

o 

o 

+  1300 

-  4600 

+      1300* 

mn 

o 

o 

+  2700 

-  4100     +  2700* 

pq 

o 

o 

+  4850 

-  3150 

+     4850* 

rs 

o 

o 

+  7080 

-  195° 

+     7080* 

COUNTERBRACING.  93 

It  is  seen  that  the  diagonals  shown  are  all  in  tension  when 
the  wind  is  from  the  right,  and  all  in  compression  when  the 
wind  is  from  the  left.  Therefore  counters  are  needed  in  all 
panels,  and  the  counters  will  all  come  in  action  whenever  the 
wind  is  from  the  left. 

(e)  Stresses  in  counterbraces.  —  It  is  evident  from  symmetry 
that  no  new  diagrams  are  needed  to  determine  the  stresses  in 
the  counterbraces.  In  fact,  the  counterbrace  in  each  panel  is 
situated  symmetrically  to  the  main  diagonal  in  another  panel, 
and  is  subject  to  exactly  equal  stresses. 

(d)  Combination  for  maxima.  —  In  combining  the  results  for 
the  greatest  stresses  in  the  various  members,  it  will  be  assumed 
that  the  greatest  snow  load  and  the  greatest  wind  load  can 
never  act  simultaneously.  For  each  member,  therefore,  the 
stress  due  to  permanent  load  is  to  be  added  to  the  snow  stress 
of  the  wind  stress,  whichever  is  greater.  Again,  in  combining 
the  tabulated  results,  we  consider  only  the  columns  headed  per- 
manent load,  snow  load,  and  wind  right ;  since  whenever  the 
wind  is  from  the  left,  the  other  system  of  diagonals  is  in  action. 
This  gives  the  results  entered  in  the  last  column. 

We  now  notice  the  following  facts : 

(1)  The  results  given  for  the  diagonal  members  are  the  true 
maximum  stresses. 

(2)  The  stress  found  for  each    diagonal  applies    also  to  the 
symmetrically  situated  counterbrace. 

(3)  For  any  other  member,   we   are  to   choose  between  the 
maximum  found  for  that  member  and  the  value  found  for  the 
symmetrically    situated    member.      Thus,    —20730  is  the  true 
maximum  stress  for  both  aj  and//;  etc.      (The  numbers  denot- 
ing true  maximum  stresses  are  marked  with  a  (*)  in  the  table.) 

It  is  seen  that  the  verticals,  with  one  exception,  may  sustain 
a  reversal  of  stress.  Thus,  jk  and  st  must  be  designed  for  a 
tension  of  2620  Ibs.,  and  also  for  a  compression  of  170  Ibs.  ; 
while  Im  and  qr  are  each  liable  to  2620  Ibs.  tension  and  to  470 
Ibs.  compression. 


CHAPTER   VI.     SIMPLE    BEAMS. 

§    i.    General  Principles. 

1 14.  Classification  of  Beams.  —  A  beam  has  been  defined  in 
Art.  79.     Beams  may  be  treated  in  two  main  classes,  the  basis 
of  classification  being  that  described  in  Art.  75.     These  two 
classes  will  be  called  simple  and  non-simple  beams  respectively. 
The  present  chapter  deals  only  with  simple  beams,  the  defini- 
tion of  which  may  be  stated  as  follows : 

A  simple  beam  is  one  so  supported  that  it  may  be  regarded 
as  a  rigid  body  in  determining  the  reactions. 

A  simple  beam  may  rest  on  two  supports  at  the  ends ;  or  it 
may  overhang  one  or  both  supports. 

A  cantilever  is  any  beam  projecting  beyond  its  supports. 
Such  a  beam  may  be  either  simple  or  non-simple. 

A  continuous  beam  is  one  resting  on  more  than  two  supports. 
Such  a  beam  is  non-simple. 

A  beam  may  be  supported  in  several  ways.  It  is  simply 
supported  at  a  point  when  it  rests  against  the  support  so  that 
the  reaction  has  a  fixed  direction.  It  is  constrained  at  a  point 
if  so  held  that  the  tangent  to  the  axis  of  the  beam  at  that 
point  must  maintain  a  fixed  direction.  If  hinged  at  a  support, 
the  reaction  may  have  any  direction.  We  shall  deal  mainly 
with  the  case  of  simple  support. 

In  what  follows  it  will  be  assumed  that  the  beam  rests  in  a 
horizontal  position,  since  this  is  the  usual  case. 

115.  External  Shear,  Resisting  Shear,  and  Shearing  Stress. 

-The  external  shear  at  any  section  of  a  beam  is  the  algebraic 

94 


GENERAL   PRINCIPLES. 


95 


sum  of  the  external  vertical  forces  acting  on  the  portion  of  the 
beam  to  the  left  of  the  section. 

The  resisting  shear  at  any  section  is  the  algebraic  sum  of  the 
internal  vertical  forces  in  the  section  acting  on  the  portion  of 
the  beam  to  the  left,  and  exerted  by  the  portion  to  the  right 
of  the  section. 

The  shearing  stress  at  any  section  is  the  stress  which  con- 
sists of  the  internal  vertical  forces  in  the  section,  exerted  by 
the  two  portions  of  the  beam  upon  each  other.  It  consists  of 
the  resisting  shear  and  the  reaction  to  it.  (See  Art.  63.) 

Let  AB  (Fig.  43)  be  a  beam  in  equilibrium  under  the  action 
of  any  external  forces.  At  any  point  in  its  length,  as  C,  con- 
ceive a  plane  to  be  passed  perpen- 
dicular  to  the  axis  of  the  beam,  and 
consider  the  portion  AC,  to  the  left 
of  the  section.  The  principles  of  equilibrium  apply  to  the  body 
AC,  and  the  external  forces  acting  upon  it  include,  besides 
those  forces  to  the  left  of  C  that  are  external  to  the  whole  bar, 
certain  forces  acting  across  the  section  at  C  that  are  internal  to 
AB,  but  external  to  AC.  (Art.  61.)  These  latter  forces  com- 
prise that  constituent  of  the  internal  stress  between  AC  and 
CB  which  is  exerted  by  CB  upon  AC. 

Represent  by  /^the  algebraic  sum  of  the  resolved  parts  in 
the  vertical  direction  of  all  forces  acting  on  AB  to  the  left  of 
the  section  at  C,  upward  forces  being  called  positive.  V  is  the 
external  shear  at  the  given  section  as  above  defined. 

Since  the  body  AC  is  in  equilibrium,  condition  (i),  Art.  58, 
requires  that  the  algebraic  sum  of  the  resolved  parts  in  the 
vertical  direction  of  all  forces  acting  on  it  must  equal  zero. 
Hence  the  forces  acting  on  AC  m  the  section  at  C  must  have  a 
vertical  component  equal  to  —  V.  This  vertical  component  is 
called  the  resisting  shear  in  the  given  section.  This  resisting 
shear  is  one  of  the  forces  of  a  stress  of  which  the  other  is  an 
equal  and  opposite  force  exerted  by  A  C  upon  CB.  This  stress 
is  called  the  shearing  stress  in  the  section,  and  is  called  positive 


96  GRAPHIC    STATICS. 

when  it  resists  a  tendency  of  AC  to  move  upward,  and  of  CB 
to  move  downward. 


1 1 6.  Bending  Moment,  Resisting  Moment,  and  Stress  Moment. 
—  The  bending  moment  at  any  section  of  a  beam  is  the  algebraic 
sum  of  the  moments  of  all  the  external  forces  acting  on  the 
portion  of  the  beam  to  the  left  of  the  section  ;  the  origin  of 
moments  being  taken  in  the  section. 

The  resisting  moment  at  any  section  is  the  algebraic  sum  of 
the  moments  of  the  internal  forces  in  the  section  acting  on  the 
portion  of  the  beam  to  the  left,  and  exerted  by  the  portion  to 
the  right  of  the  section ;  the  origin  of  moments  being  the  same 
as  for  bending  moment. 

The  stress  moment  or  moment  of  internal  stress  at  any  section 
consists  of  the  two  equal  and  opposite  moments  of  the  forces 
exerted  across  the  section  by  the  two  portions  of  the  beam 
upon  each  other. 

Referring  again  to  Fig.  43,  let  us  analyze  further  the  forces  in 
the  section  at  C.  Applying  to  the  body  AC  the  second  condition 
of  equilibrium  ((2)  of  Art.  58),  and  taking  an  origin  at  a  point 
in  the  section,  we  see  that  the  algebraic  sum  of  the  moments 
of  all  the  external  forces  acting  on  the  beam  to  the  left  of  the 
section  plus  the  sum  of  the  moments  of  the  internal  forces  act- 
ing on  AC  in  the  section  must  equal  zero.  The  former  sum  is 
defined  as  the  bending  moment  at  the  given  section.  Represent 
it  by  M.  The  latter  sum  is  defined  as  the  resisting  moment  at 
the  section,  and  must  be  equal  to  —My  by  the  above  principle. 

We  have  thus  far  referred  only  to  the  internal  forces  exerted 
by  CB  upon  AC]  but  evidently  the  equal  and  opposite  forces 
exerted  by  A  C  upon  CB  have  a  moment  numerically  equal  to 
M.  The  equal  and  opposite  moments  of  the  equal  and  opposite 
forces  of  the  stress  in  the  section  together  constitute  the  stress 
moment  in  the  section. 

If  the  external  forces  applied  to  the  beam  are  all  vertical,  the 
value  of  M  will  be  the  same  at  whatever  point  of  the  section 


GENERAL   PRINCIPLES.  97 

the  origin  is  taken  ;  since  the  arm  of  each  force  will  be  the  same 
for  all  origins  in  the  same  vertical  line.  If  the  loads  and  reactions 
are  not  all  vertical,  the  value  of  M  will  generally  depend  upon 
what  point  in  the  section  is  taken  as  the  origin  of  moments. 

117.  Curves  of  Shear  and  Bending  Moment. — The  curve  of 
shear  for  a  beam  is  a  curve  whose  abscissas  are  parallel  to  the 
axis  of  the  beam,  and  whose  ordinate  at  any  point  represents 
the  external  shear  at  the  corresponding  section  of  the  beam. 

Let  AB  (Fig.  44)  "represent  a  beam  loaded  in  any  manner,  and 
let  A'B1  be  taken  parallel  to  AB.  At  every  point  of  A'B' 
suppose  an  ordinate  drawn  whose  length  shall  represent  the 
external  shear  at  the  corresponding  B 

section  of  AB.  The  line  ab,  join- 
ing the  extremities  of  all  these  ordi- 
nates,  is  the  shear  curve.  Positive 
values  of  the  shear  may  be  repre- 
sented by  ordinates  drawn  upward 
from  A'B',  and  negative  values  by 
ordinates  drawn  downward.  (Instead  of  drawing  A'B'  parallel 
to  the  beam,  it  may  be  any  other  straight  line  whose  extremi- 
ties are  in  vertical  lines  through  A  and  B.) 

The  curve  of  bending  moment  for  a  beam  is  a  curve  whose 
abscissas  are  parallel  to  the  axis  of  the  beam,  and  whose  ordinate 
at  any  point  represents  the  bending  moment  at  the  correspond- 
ing section  of  the  beam.  Thus  in  Fig.  44,  A"C"B"  may  repre- 
sent the  bending  moment  curve  for  the  beam  AB.  (Evidently 
A"B"  might  be  inclined  to  the  direction  of  AB,  without  destroy- 
ing the  meaning  of  the  curve.)  Positive  and  negative  values  of 
the  bending  moment  will  be  distinguished  by  drawing  the 
ordinates  representing  the  former  upward  and  those  represent- 
ing the  latter  downward  from  the  line  of  reference  A"B". 

1 1 8.  Moment  Curve  a  Funicular  Polygon.  —  If  the  loads  and 
reactions  upon  the  beam  are  all  vertical,  every  funicular  poly- 
gon for  these  forces  taken  consecutively  along  the  beam,  is  a 


98 


GRAPHIC    STATICS. 


curve  of  bending  moments.  Thus,  let  MN  (Fig.  45)  represent 
a  beam  under  vertical  loads,  supported  at  the  ends  by  vertical 
reactions.  Draw  a  funicular  polygon  for  the  loads  and  reactions 
as  shown. 


M 


«|Z) 
t 


Fig.  45 

Now,  by  definition,  the  bending  moment  at  any  section  is 
equal  to  the  moment  of  the  resultant  of  all  external  forces  act- 
ing on  the  beam  to  the  left  of  the  section,  the  origin  of 
moments  being  taken  in  the  section.  By  Art.  56,  this  moment 
can  be  found  by  drawing  through  the  section  a  vertical  line  and 
finding  the  distance  intercepted  on  it  by  the  two  strings  corre- 
sponding to  the  resultant  mentioned  ;  the  product  of  this  inter- 
cept by  the  pole  distance  is  the  required  moment.  Hence,  if 
oe  (Fig.  45)  is  taken  as  axis  of  abscissas,  the  broken  line  made 
up  of  oa,  ob,  oc,  od,  is  a  "curve  of  bending  moments." 

119.  Design    of    Beams. — The   principles   involved   in   the 
design  of  beams  will  not  be  here  fully  discussed.     Every  prob- 
lem in  design  involves  the  determination  of  shears  and  bend- 
ing moments  throughout  the  beam ;  and  the  graphic  methods 
of  determining  these  will  alone  be  considered  in  the  following 
articles. 

§  2.  Beam  Sustaining  Fixed  Loads. 

120.  Shear  Curve  for  Beam  Supported  at  Ends.  —  Let  MN 
(Fig.  45)  represent  a  beam  supported  at  the  ends  and  sustain- 
ing loads  as  shown.     Draw  the  force  polygon  ABCD,  and  with 
pole    O   draw  the   funicular   polygon.     The  closing  line  is  oe 


BEAM    SUSTAINING   FIXED    LOADS.  99 

(marked  also  M"N"),  and  OE  drawn  parallel  to  M"N"  fixes  E, 
thus  determining  DE,  EA,  the  reactions  at  the  supports. 

Take  M'N'  as  the  axis  of  abscissas  for  the  shear  curve. 

The  shear  at  every  section  can  be  at  once  taken  from  the 
force  polygon.  For,  remembering  the  definition  of  external 
shear  (Art.  1 15)  we  have  : 

Shear  at  any  section  between  J/and  P  is  EA  (positive). 

Shear  at  any  section  between  P  and  Q  is  EB  (positive). 

Shear  at  any  section  between  Q  and  R  is  EC  (negative). 

Shear  at  any  section  between  R  and  N  is  ED  (negative). 

Hence  the  shear  curve  is  the  broken  line  drawn  in  the 
figure. 

121.  Moment   Curve  for   Beam   Supported  at  Ends. — As  in 

Art.  1 1 8,  it  is  seen  that  the  funicular  polygon  already  drawn 
(Fig.  45)  is  a  bending  moment  curve  for  the  given  forces.  For 
the  bending  moment  at  any  section  of  the  beam  is  equal  to  the 
corresponding  ordinate  of  this  polygon,  multiplied  by  the  pole 
distance. 

For  simplicity,  it  will  be  well  always  to  choose  the  pole  so 
that  the  pole  distance  represents  some  simple  number  of  force- 
units. 

The  sign  of  the  bending  moment  is  readily  seen  to  be  nega- 
tive everywhere,  according  to  the  convention  already  adopted 
(Art.  47). 

122.  Shear   and   Moment   Curves  for  Overhanging  Beam. — 

Consider  a  beam  such  as  shown  in  Fig.  46,  supported  at  Q  and 
T,  and  sustaining  loads  at  M,  P,  R,  S,  and  N. 

This  case  may  be  treated  just  as  the  preceding,  care  being 
exercised  to  take  all  the  external  forces  (loads  and  reactions)  in 
order  around  the  beam. 

The  construction  is  shown  in  Fig.  46.  First,  the  reactions  at 
Q  and  T  are  found  as  in  the  preceding  case,  by  drawing  the 
funicular  polygon,  finding  the  closing  line,  and  drawing  OG 
parallel  to  it.  The  reactions  are  FG  and  GA.  Then  the  value 


100 


GRAPHIC    STATICS. 


of  the  external  shear  can  be  found  for  any  section  from  the 
definition,  and  is  always  given  in  magnitude  and  sign  by  a  cer- 
tain portion  of  the  force  polygon  ABCDEFGA.  The  resulting 
curve  is  shown  in  the  figure,  M'N'  being  the  line  of  reference. 


Fig.  46 

Similarly,  from  the  definition  of  bending  moment,  and  the 
principle  of  Art.  56,  the  bending  moment,  at  any  section  is 
equal  to  the  ordinate  of  the  funicular  polygon  multiplied  by  the 
pole  distance.  The  polygon  is  shaded  in  the  figure  to  indicate 
the  ordinates  in  question.  It  is  seen  that  the  bending  moment 
is  positive  at  every  section  of  the  beam. 

123.  Distributed  Loads.  —  In  all  the  cases  thus  far  discussed, 
the  loads  applied  to  the  beam  have  been  considered  as  concen- 
trated at  a  finite  number  of  points.  That  is,  it  has  been 
assumed  that  a  finite  load  is  applied  to  the  beam  at  a  point. 
Such  a  condition  cannot  strictly  be  realized,  every  load  being 
in  fact  distributed  along  a  small  length  of  the  beam.  If  the 
length  along  which  a  load  is  distributed  is  very  small,  no  im- 
portant error  results  from  considering  it  as  applied  at  a  point. 

In  certain  cases  a  beam  may  have  to  sustain  a  load  which  is 
distributed  over  a  considerable  part  of  its  length,  or  over  the 
whole  length.  Such  a  load  may  be  treated  graphically  with 
sufficient  correctness  by  dividing  the  length  into  parts,  and 


BEAM    SUSTAINING    MOVING 


assuming  the  whole  load  on  any  part  to  be  concentrated  at  a 
point.  The  smaller  these  parts,  the  more  nearly  correct  will 
the  results  be. 

It  may  be  remarked  by  way  of  comparison  that  while  algebraic 
methods  are  most  readily  applicable  to  the  case  of  distributed 
loads,  the  reverse  is  true  of  graphic  methods,  which  are  most 
easily  applied  in  the  very  cases  in  which  analytic  methods 
become  most  perplexing. 

124.  Design  of  Beam  with  Fixed  Loads.  —  The  above  exam- 
ples are  sufficient  to  explain  the  method  of  treating  any  simple 
beam  under  fixed  loads.  Under  such  loading  the  shear  and 
bending  moment  at  each  section  of  the  beam  remain  unchanged 
in  value,  and  no  further  discussion  is  necessary  as  a  preliminary 
to  the  design  of  the  beam.  We  proceed  next  to  the  case  of 
beams  with  moving  loads. 

§  3.    Beam  Sustaining  Moving  Loads. 

12$.  Curves  of  Maximum  Shear  and  Moment.  —  When  a  beam 
sustains  moving  loads,  the  shear  and  moment  at  any  section  do 
not  remain  constant  for  all  positions  of  the  loads.  In  such  a 
case  it  is  the  greatest  shear  or  moment  in  each  section  that  is 
to  be  used  in  designing  the  beam. 

A  curve  of  maximum  shear  is  a  curve  of  which  the  ordinate 
at  each  point  represents  the  greatest  possible  value  of  the  shear 
at  the  corresponding  section  of  the  beam  for  any  position  of  the 
loads. 

A  curve  of  maximum  moment  is  a  curve  whose  ordinate  at 
each  point  represents  the  greatest  possible  value  of  the  bending 
moment  at  the  corresponding  section  of  the  beam  for  any  posi- 
tion of  the  loads. 

In  the  following  articles  will  be  explained  a  method  of  deter- 
mining any  number  of  points  of  the  curves  just  defined,  in  the 
case  of  a  simple  beam  supported  at  the  ends.  The  moving 


•(02  GRAPHIC    STATICS. 

load  will  be  taken  to  consist  of  a  series  of  concentrated  loads 
with  lines  of  actions  at  fixed  distances  apart.  An  example  of 
such  a  load  is  the  weight  of  a  locomotive  and  train ;  the 
points  of  application  of  the  loads  being  always  under  the  several 
wheels. 

126.  Position  of  Loads  for  Greatest  Shear  at  Any  Section.— 
In  Fig.  47  (PL  III)  let  the  vertical  lines  ab,  be,  etc.,  represent 
the  lines  of  action  of  the  moving  loads  in  their  true  relative 
positions  ;  the  magnitudes  of  the  loads  being  shown  in  the  force 
polygon  or  "load  line"  ABCD  .  .  .  Let  XY  represent  the 
length  of  the  beam,  and  let  it  be  required  to  determine  the 
position  of  the  moving  loads  that  will  cause  the  maximum  posi- 
tive shear  at  any  section,  as  at  that  marked  J. 

First,  consider  the  effect  of  a  single  load  in  any  position.  It 
is  seen  that  any  load  to  the  right  of  J  produces  at  that  point  a 
positive  shear.  For,  by  definition,  the  external  shear  is  equal 
to  the  algebraic  sum  of  all  vertical  forces  to  the  left  of  the 
section,  upward  forces  being  reckoned  positive.  Now  a  load 
to  the  right  of  J  causes  no  forces  to  act  at  the  left  of  this 
section,  except  an  upward  reaction  at  X.  On  the  other  hand, 
a  load  to  the  left  of  J  produces  negative  shear  at  that  point. 
For  the  shear  due  to  it  is  equal  to  the  reaction  at  X  minus  the 
load  itself,  which  will  always  be  negative,  since  the  load  is 
greater  than  the  reaction  due  to  it.  (It  is  also  to  be  noticed 
that  the  shear  due  to  a  load  at  the  left  of  the  section  is  equal 
numerically  to  the  reaction  it  produces  at  K) 

Again,  the  shear  due  to  any  load  is  greater  in  magnitude,  the 
nearer  the  load  is  to  the  section  considered.  For,  as  a  load  on 
JY  approaches  J,  the  reaction  at  X  increases  ;  and  as  a  load 
on  XJ  approaches  _/,  the  reaction  at  Y  increases. 

Two  principles  are  thus  reached  to  guide  in  assigning  the 
position  of  the  loads  that  will  produce  the  greatest  positive 
shear  at  any  section  :  (i)  The  segment  of  the  beam  to  the  left 
of  the  section  should  be  without  load  and  that  to  the  right  fully 


BEAM    SUSTAINING   MOVING    LOADS.  IO3 

loaded  ;  and  (2)  the  loads  on  the  loaded  segment  should  be  as 
near  the  section  as  possible. 

It  would  seem,  therefore,  that  the  greatest  positive  shear  at 
the  point  J  will  occur  when  the  loads  are  brought  on  the  beam 
from  the  right  to  such  a  position  that  the  foremost  load  in  the 
series  is  just  at  the  right  of  J.  And,  in  general,  this  will  be 
true,  unless  the  foremost  load  is  small  in  comparison  with  the 
whole  load  on  the  beam,  or  unless  the  distance  between  the  first 
and  second  loads  is  considerable.  In  such  cases  it  may  be  that 
a  greater  shear  will  result  when  the  second  load  is  brought  up 
to  the  section.  For,  suppose  the  first  load  to  be  just  at  the 
right  of  J,  and  consider  the  effect  of  moving  the  whole  series 
of  loads  to  the  left.  As  the  first  load  passes  the  section,  it 
produces  a  diminution  of  the  shear  equal  in  amount  to  the  load. 
On  the  other  hand,  as  the  loads  continue  to  move  to  the  left, 
the  effect  of  every  load  in  producing  reaction  at  X  increases, 
while  no  further  decrease  in  the  shear  at  J  occurs  until  the 
second  load  passes  the  section.  Now,  it  may  be  that  the  net 
result  of  bringing  the  second  load  up  to  J  will  be  to  increase 
the  shear  at  that  section.  If  the  first  two  or  more  loads  are 
very  small,  it  may  be  that  the  third  or  fourth  load  should  be 
brought  to  the  section  to  produce  the  greatest  shear.  Usually, 
however,  it  will  be  necessary  to  try  only  two  (or  at  most  three) 
positions. 

Instead  of  resorting  to  trials  to  determine  for  which  position 
the  shear  is  greatest,  we  may  apply  a  simple  rule  which  will 
now  be  deduced. 

Let  PI  =  magnitude  of  foremost  load,  P^—  magnitude  of  sec- 
ond load,  etc.,  W  being  the  total  load  on  the  beam.  Let  /= 
total  span,  and  m  =  distance  between  PI  and  P*  Let  x=  dis- 
tance from  Y  to  the  center  of  gravity  of  W  when  Pv  is  at  a 
given  section.  Let  us  compute  the  shear  when  Pl  is  just  at 
the  right  of  the  section,  and  then  determine  the  effect  of  moving 
all  loads  to  the  left,  until  the  second  load  comes  to  the  section. 

When  P  is  at  the  right  of  the  section,  the  shear  is  equal  to 


IO4 


GRAPHIC    STATICS. 


the  reaction  at  X,  say  R.     Then  (calling  V  the  external  shear) 
we  have 


. 

If  now  the  loads  be  moved  until  P2  comes  to  a  point  just 
at  the  right  of  the  section,  the  reaction   due  to    W  becomes 

W  \x+m\  and  the  shear  at  the  section  becomes 


The  increase  of  the  shear  is,  therefore, 

„       Wx_  Wm      D 
~~ 


This  increase  is  plus  or  minus  according  as  —  —  is  greater  or 

P  / 

less  than  Pl  ;  that  is,  as  IV  is  greater  or  less  than  —  -.      Hence 

m 

the  following  rule  : 

The  maxim  tun  positive  shear  in  any  section  of  the  beam  occurs 
when  the  foremost  load  is  infinitely  near  the  section,  provided  W 

PI  PI 

is  not  greater  than  —  .     If  W  is  greater  than  —  ^,  the  greatest 
m  m 

shear  will  occur  wJien  some  succeeding  load  is  at  the  section. 

In  the  above  discussion  it  has  been  assumed  that  in  bringing 
P2  up  to  the  section  no  additional  loads  are  brought  upon  the 
beam.  If  this  assumption  is  not  true,  let  W  be  the  new  load 
brought  on  the  beam,  and  x'  the  distance  of  its  center  of 
gravity  from  the  right  support  when  P.t  is  at  the  section.  Then 
the  shear  corresponding  to  this  position  is 


and  the  increase  of  shear  due  to  the  assumed  change  in  the 
position  of  the  load  is 

Wm  ,   W1*'      v 
-      +  —        ft 


BEAM    SUSTAINING    MOVING   LOADS. 


105 


Hence,  in   the  statement   of  the  above   rule,  we  have   only  to 

substitute    Wm+W'x'  for    Wm\   or,    JF+-2-*    for    W.     The 

m 

additional  load  W  may  be  neglected  except  when  the  condition 

P  I 

W  =  — -  is  nearly  satisfied;  for  the  term   W*x?  will  always  be 

m 

small  in  comparison  with   Wm. 

The  application  of  this  rule  is  very  easy,  and  will  save  much 
labor  in  the  graphic  construction  of  the  shear  curve. 

If  it  is  found  that  the  first  load  should  be  past  the  section 
for  the  position  of  greatest  shear,  we  may  determine  whether 
the  second  or  the  third  should  be  at  the  section  by  an  exactly 
similar  method.  We  have  only  to  apply  the  above  rule,  substi- 
tuting P.2  for  />!,  and  for  m  the  distance  between  P  and  P3. 

127.  Determination  of  Maximum  Shear.  — The  shear  at  any 
section  due  to  any  position  of  the  moving  loads  can  readily  be 
determined  from  the  force  and  funicular  polygons,  by  a  method 
which  will  now  be  shown. 

Draw  in  succession  the  lines  of  action  of  the  loads  at  their 
proper  distances  apart,  as  in  Fig.  47  (PL  III).  Draw  the  load- 
line  ABCD  .  .  .,  and  choosing  a  pole  O,  draw  the  funicular 
polygon.  The  same  space  diagram  may  be  used-for  any  position 
of  the  moving  loads  ;  for,  instead  of  moving  the  loads  in  either 
direction,  we  may  assume  the  loads  fixed  and  move  the  beam 
in  the  opposite  direction. 

By  applying  the  rule  deduced  in  Art.  126,  it  may  be  shown 
that  there  is  a  point  Z  in  the  beam,  such  that  for  any  section 
to  the  right  of  Z  the  foremost  load  should  be  at  the  section  in 
order  that  the  shear  may  be  a  maximum  ;  while  for  any  section 
to  the  left  of  Z  the  second  load  must  be  brought  to  the  section. 
The  position  of  this  point  is  readily  determined,  as  follows  : 

In  the  case  shown  in  Fig.  47,  we  have 

/-64ft.,  01  =  8.1  ft.,  P1  =  Sooo  Ibs.,  ^=^1x8000  =  63200. 

;;/       8.1 

Hence,  as  the  load  moves  from  right  to  left,   W  is  less  than 


I06  GRAPHIC    STATICS. 

P  I 

—   until  the  whole  load  on  the  beam  reaches  65200  Ibs.     This 

m 

occurs  when  the  fifth  load  is  at  Y.  Hence  the  point  Z  is 
located  at  a  distance  from  Y  equal  to  the  distance  between  the 
lines  of  action  of  the  first  and  fifth  loads. 

For  any  point  to  the  left  of  Z,  the  greatest  shear  will  prob- 
ably occur  when  the  second  load  is  at  the  section.  To  test 
whether  in  any  case  the  third  load  should  be  at  the  section,  we 
apply  the  same  principle,  as  follows  :  The  second  load  is  1 5000 
Ibs.  ;  the  distance  between  the  second  and  third  loads  is  5.8  ft. 
Hence,  for  greatest  shear,  the  third  load  should  not  be  at 
the  section  unless  the  whole  load  on  the  beam  is  at  least 

—  x  15000  Ibs.  or  165500  Ibs.     But  it  is  easily  seen   that  the 

total  load  can  never  be  so  great. 

We  may  now  explain  the  construction  for  finding  the  great- 
est shear  at  any  section.  Consider  a  section  at  the  left  of  Z, 
as  at  the  point  marked  J.  From  the  above  reasoning  it  is  evi- 
dent that  the  second  load  must  be  brought  close  to  the  section. 
In  Fig.  47,  SjSj  is  drawn  to  represent  the  beam,  its  position 
being  such  that  the  point  J  is  infinitely  near  be,  the  line  of 
action  of  the  second  load.  Through  the  extremities  of  the 
beam  SjSj  draw  vertical  lines  to  represent  the  action-lines  of 
the  reactions  at  the  supports.  The  portion  AB'  of  the  load 
line  represents  the  loads  now  on  the  beam,  and  the  correspond- 
ing extreme  strings  of  the  funicular  polygon  are  oa  and  ob'.  If 
now  we  consider  the  funicular  polygon  for  the  loads  and  reac- 
tions acting  on  the  beam  in  the  supposed  position,  we  see  that 
the  closing  side  is  found  by  joining  the  points  in  which  the 
action-lines  of  the  end  reactions  intersect  oa  and  ob'  respec- 
tively. This  is  marked  of  in  the  figure.  Now  draw  in  the 
force  diagram  a  ray  parallel  to  this  closing  line,  and  let./*  be  its 
point  of  intersection  with  the  load  line ;  then  B'J1  andJ'A  are 
the  two  reactions.  The  shear  at  the  point  J  is  evidently  the 
algebraic  sum  of  the  reaction  J'A  and  the  load  AB ;  hence  it 


BEAM    SUSTAINING    MOVING    LOADS. 


107 


is  represented  byj'fi.  In  Fig.  47  let  X,Y,  be  drawn  to  repre- 
sent the  length  of  the  beam,  and  let  the  ordinates  of  the  shear 
curve  be  drawn  from  it.  Then  from  J  we  lay  off  an  ordinate 
equal  toj'£>.  The  whole  curve  of  maximum  shear  is  shown  in 
the  figure,  but  the  construction  for  finding  it  is  given  for  only 
one  point. 

128.  Shear  Curve  for  Combined  Fixed  and  Moving  Loads.  - 
Let  the  beam  sustain  a  fixed  load  of  25000  Ibs.  uniformly  dis- 
tributed along  the  beam.     The  shear  close  to  the  left  support 
due  to  this  load  is  equal  to  the  reaction,  or  12500  Ibs.  ;  and 
decreases  as  we  pass  to  the  right  by  ^f^-^-  Ibs.  for  each  foot. 
At  the  middle  of  the  beam  the  shear  is  zero,  and  at  the  right 
support  it  is  —12500  Ibs.     Hence  the  shear  curve  is  a  straight 
line,  and  may  be  drawn  as  follows  :  From  X,  lay  off  an  ordinate 
downward    representing   12500  Ibs,,  and  from    Ys  an  ordinate 
upward  representing  12500  Ibs.;  the  straight  line  joining  the 
extremities  of  these  ordinates  is  the  shear  curve  for  the  fixed 
load.      Positive    shears   are   laid   off    downward    and   negative 
shears  upward  for  the  reason  that,  if  this  be  done,  the  greatest 
positive  shear  at  any  point  due  to  fixed,  and  moving  loads  is 
represented  by  the  total  ordinate  measured  between   the   shear 
curves  for  fixed  loads  and  for  moving  loads.     It  is  seen  that  at 
a  certain  point  somewhere  to  the  right   of  the  center  of  the 
beam  this  greatest  shear  is  zero,  and  for  all  sections  to  the 
right  of  this  point  it  is  negative.     This  point  is  determined  by 
the  intersection  of  the  two  shear  curves. 

1 29.  Position    of     Loads   for   Greatest  Bending    Moment.  — 
Referring  to  the  beam  XY  shown  in  Fig.  47  (PL  III),  consider 
the  bending  moment  at  any  section   (asy)  due  to  a  load  any- 
where on  the  beam.     First,  suppose  the  load  is  at  the  right  of 
the  section.     In  this  case  the  only  force  brought  upon  the  por- 
tion of  the  beam  to  the  left  of  the  section  is  a  reaction  at  X. 
By  definition  (Art.  116)  the  bending  moment  aty  is  the  moment 
of  this  reaction  about  an  origin  aty,  and  is  negative.     Moreover, 


I08  GRAPHIC    STATICS. 

since  the  reaction  will  become  greater  as  the  load  moves  toward 
the  left,  the  greatest  bending  moment  due  to  a  load  at  the  right 
of  the  section  will  occur  when  the  load  is  as  near  the  section  as 
possible.  Next  consider  a  load  at  the  left  of  the  section.  The 
bending  moment  due  to  it  is  equal  to  the  moment  of  the  result- 
ant of  the  load  and  the  left  reaction.  But  this  resultant  is 
equal  and  opposite  to  the  reaction  at  the  right  support,  and  has 
the  same  line  of  action  (because  the  load  and  the  two  reactions 
due  to  it  are  in  equilibrium)  ;  hence  the  bending  moment  is  the 
negative  of  the  moment  of  the  right  reaction  about  an  origin 
in  the  section.  The  bending  moment  is  therefore  negative, 
and  is  greatest  when  the  right  reaction  is  greatest.  Hence,  a 
load  at  the  left  of  the  section  produces  its  greatest  bending 
moment  when  the  load  is  as  near  as  possible  to  the  section. 

We  are  therefore  led  to  the  following  general  principles  for  a 
simple  beam  supported  at  the  ends  : 

(1)  The  bending  moment  at  every  section  is  negative  for  all 
positions  of  the  moving  loads. 

(2)  The    negative   bending   moment  at   any  section  has    its 
greatest  value  when  the  whole  beam  is  loaded  as  completely  as 
possible,  and  the  heaviest  loads  are  near  the  section. 

These  principles  serve  as  a  general  guide,  but  unless  the 
moving  loads  are  equidistant  and  equal  in  magnitude  it  will  be 
necessary  to  try  several  positions  before  the  position  for  great- 
est bending  moment  will  be  known.  Usually  one  of  the 
heaviest  loads  should  be  directly  at  the  section. 

Instead  of  resorting  to  repeated  trials,  the  position  of  loads 
giving  greatest  bending  moment  at  any  section  may  be  found 
by  means  of  a  simple  rule  which  will  now  be  deduced. 

Let  /=  length  of  beam  between  supports;  W=  total  load  on 
beam  ;  x  =  distance  of  center  of  gravity  of  W  from  right  sup- 
port ;  W7!  — load  on  beam  to  left  of  given  section;  ,r=distance 
of  center  of  gravity  of  W\  from  section;  l±  =  distance  of  sec- 
tion from  left  support ;  R  =  reaction  at  left  support  ;  M — 
bending  moment  at  the  section.  Then  we  have 


BEAM    SUSTAINING    IVIOVING   LOADS. 
Wr 


If  the  loads  all  move  an  infinitesimal  distance  to  the  left, 
then  x  and  ,TI  receive  equal  infinitesimal  increments,  and  the 
increment  of  M  is 


dM=       -dx-  Wjlxi  =      -    -  W\  dx 

(since   dxl  =  dx).     Now   if   M  is    a   maximum,  we    must    have 

—  =  o;  that  is,  -^-  ^  =  o;  or,    }V±=™-.    Hence  the  follow- 
dx  /  A         / 

ing  rule  : 

W7/^#  //#•  bending  moment  at  any  section  of  the  beam  has  its 
greatest  vahie,  tJie  loads  on  the  tivo  segments  of  tJie  beam  are  to 
each  other  as  the  lengths  of  the  segments. 

In  case  of  concentrated  loads,  the  condition  just  stated  can 
generally  not  be  exactly  satisfied  except  for  certain  sections  of 
the  beam.  It  will  usually  be  found  that  the  position  most 
nearly  fulfilling  the  condition  is  that  in  which  some  heavy  load 
is  just  at  the  section.  This  will  be  illustrated  in  the  next 
article. 

[NOTE.  —  The  above  reasoning  applies  only  to  the  case  of  concentrated  loads,  and 
is  not  rigorous  for  this  case,  if,  in  the  position  of  maximum  moment  for  any  section, 

a  load  is  at  the  section  ;  for  -  l  will  not  then  be  zero,  as  assumed  above.     In  such 

dM 
a  case  —  —  is  discontinuous,  and  its  value  changes  sign  as  the  loads  pass  through  the 

dX  rz^.  JJT 

position   for  which  —  i  --  =  o.     It  is  still  true  that  this  condition  is  satisfied  in 
/!          / 

the  position  of  maximum  moment,  provided  the  load  at  the  section  is  regarded  as 
divided  in  some  certain  ratio  between  the  two  segments  of  the  beam.] 

130.  Determination  of  Bending  Moments.  —  The  application 
of  the  above  principles  and  the  method  of  determining  the 
bending  moment  at  any  section  of  the  beam  will  now  be  shown 
for  the  case  represented  in  Fig.  47  (PL  III).  Let  the  greatest 
bending  moment  be  found  for  the  section  J.  From  the  general 


1 10  GRAPHIC    STATICS. 

principles  deduced  in  Art.  129,  it  is  probable  that  the  loads 
nearest  the  section  should  be  the  third  and  fourth  in  the  series. 
Also,  since  the  point  J  divides  the  span  into  segments  of  16  ft. 
and  48  ft.,  the  load  on  the  left  segment  should  be  one-fourth 
the  total  load  on  the  beam.  Now  it  is  readily  found  that  when 
the  load  CD  is  just  at  the  right  of  the  section,  the  whole  load 
on  the  beam  is  112,000  Ibs.,  while  the  load  on  the  left  segment 
is  23,000  Ibs.,  which  is  less  than  one-fourth  of  112,000.  And 
when  the  load  CD  is  just  at  the  left  of  the  section,  the  load  on 
the  left  segment  is  38,000  Ibs.,  which  is  greater  than  one-fourth 
of  112,000.  Hence  the  bending  moment  is  a  maximum  when 
the  load  CD  is  just  at  the  section. 

In  Fig.  47,  MjMj  represents  the  beam  for  this  position  of  the 
loads,  and  the  closing  line  of  the  funicular  polygon  is  marked 
of.  Using  the  principle  of  Art.  56,  we  find  the  bending  mo- 
ment at  the  given  section  by  measuring  the  distance  intercepted 
on  cd  by  the  strings  od  and  of  and  multiplying  it  by  the  pole 
distance  in  the  force  diagram. 

Let  Xm  Ym  (Fig.  47)  be  the  axis  of  abscissas  for  the  curve  of 
maximum  moments.  From  the  point  J  draw  an  ordinate  equal 
to  the  intercept  just  found ;  this  locates  a  point  of  the  required 
moment  curve. 

Other  points  may  be  determined  in  the  same  manner.  The 
curve  is  shown  in  the  figure,  but  the  construction  is  not  given 
for  any  section  except  that  at  J. 

It  must  be  remembered  that  each  ordinate  is  to  be  multiplied 
by  the  pole  distance.  Hence  it  will  be  convenient  to  choose 
the  pole  distance  equal  to  some  round  number  of  force  units. 

131.  Moment  Curve  for  Fixed  Loads.  — The  greatest  bending 
moment  due  to  moving  loads  must  be  combined  with  the  bend- 
ing moment  due  to  fixed  loads.  If  the  fixed  load  is  uniformly 
distributed,  as  already  assumed  in  the  computation  of  shear, 
it  may  be  divided  into  parts,  each  assumed  concentrated  at 
its  center  of  gravity,  arid  a  funicular  polygon  drawn,  using  the 


BEAM    SUSTAINING  MOVING    LOADS.  Iir 

same  pole  distance  employed  in  the  force  diagram  for  moving 
loads.  The  ordinates  of  this  funicular  polygon  may  be  laid  off 
upward  from  the  line  XmYm,  and  their  ends  joined  to  form  a 
continuous  curve.  The  total  ordinate  from  this  curve  to  that 
already  drawn  for  moving  loads  represents  the  true  greatest 
bending  moment  at  the  corresponding  section  of  the  beam. 
The  curve  is  shown  in  the  figure,  but  the  construction  is 
omitted,  since  it  involves  no  new  principle. 

It  is  to  be  remembered  that  the  bending  moment  found  for 
any  section  is  a  possible  value  for  the  other  section  equally 
distant  from  the  center  of  the  beam,  since  the  train  may  be 
headed  in  the  opposite  direction,  and  the  same  construction 
made,  viewing  the  beam  from  the  other  side.  The  same  state- 
ment holds  as  to  shears. 

132.  Design  of  Beam  sustaining  Moving  Loads.  —  In  designing 
a  beam  to  sustain  moving  loads,  the  greatest  shear  and  bending 
moment  that  can  come  upon  it  for  any  position  of  the  loads  must 
be  known  for  every  section.      The  methods  above  given  are 
sufficient  for  the  determination    of  these  quantities ;  and  the 
problem  of  designing  the  beam  will  not  be  here  further  discussed. 

133.  Plate  Girders  and  Lattice  Girders.  — A  plate  girder  is  a 
beam  built  up  of  rolled  plates  and  angle-irons  riveted  together, 
the  cross-section   being  as   shown   in   Fig.  48.     If 

latticing  is  substituted  for  the  plate,  the  beam  be- 
comes a  lattice  girder.  Railway  bridges  of  spans 
under  100  ft.  frequently  employ  either  rolled  beams, 
plate  girders,  or  lattice  girders.  (See  Cooper's 
"  Specifications  for  Iron  and  Steel  Railroad 
Bridges.")  The  methods  given  in  the  preceding 
articles  are  especially  useful  in  designing  this  class 
of  bridges. 

The  student  should  make  the  complete  construe- 
tion  for  determining  the  curves  of  maximum  shear  and  bending 
moment  for  a  beam  designed  to  carry  the  series  of  moving  loads 
shown  in  Fig.  47. 


CHAPTER    VII.      TRUSSES    SUSTAINING    MOVING 

LOADS. 

§   i.    Bridge  Loads. 

134.  General  Statement.  — The  most  important  class  of  trusses 
sustaining   moving  loads   is   that  of  bridge  trusses.     The  two 
main  classes  of  bridges  are  highway  bridges  and  railway  bridges. 
The  forms  of  trusses  most  commonly  used  differ  for  the  two 
classes,  as  do  also  the  amount  and  distribution  of  loads. 

Before  the  design  can  be  correctly  made,  the  weights  of  the 
trusses  themselves  must  be  known.  Since  these  weights  de- 
pend upon  the  dimensions  of  the  truss  members,  they  cannot 
be  known  with  certainty  until  the  design  is  completed.  The 
remarks  made  in  Art.  82  regarding  the  design  of  roof  trusses 
are  here  applicable. 

In  the  following  articles  we  shall  give  data  available  for 
preliminary  estimates  of  truss  weights. 

As  stated  in  Art.  133,  railway  bridges  of  short  span  are  fre- 
quently supported  by  rolled  or  built  beams.  When  the  span  is 
longer  than  100  ft.,  trusses  should  be  used.  (Cooper's  <<r  Speci- 
fications.") 

135.  Loads  on  Highway   Bridges.  —  (i)  Permanent  load. — 
The  permanent  load  sustained   by  a  highway  bridge  truss   in- 
cludes the  weight  of  the  truss  itself,  of  the  lateral  or  "sway" 
bracing,  of  the  floor  and  the  beams  and  stringers  supporting  it. 
These  weights  are  all  subject  to  much  variation,  but,  for  pur- 
poses of  preliminary  design,  the  following  formula,  taken  from 
Merriman's  "Roofs  and  Bridges,"  may  be  used. 


BRIDGE    LOADS.  Ir^ 

Let  w  —  weight  of  bridge  in  pounds  per  linear  foot ;  /=span 
in  feet  ;  b  =  width  in  feet.  Then 

w  =  140+  12  b  +  o.2  bl—o.4  I. 

(2)  Snow  load.  —  The  weight  of   snow  may  be  taken   as  in 
case  of  roof  trusses    (Art.    84).     The  values  there  given    are 
probably  in  excess  of  those  ordinarily  employed  in  practice. 

(3)  Wind  load.  —  The  pressure  of  wind  striking  the  bridge 
laterally  is   resisted  by  the  chord  members   together  with  the 
lateral  bracing.     These  constitute  horizontal  trusses,  in  which 
the  stresses  are  to  be  found  in  the  same  way  as  for  the  main 
trusses  of  the  bridge.     As  the  determination  of  wind  stresses 
requires  the  use  of  no  special  methods  or  principles,  they  will 
not    be    here    considered.     The    student   is   referred   to   Burr's 
"  Stresses    in    Roofs    and    Bridges,"    Merriman's    "  Roofs    and 
Bridges,"  and  other  available  works  for  a  complete  discussion 
of  wind  pressure  and  its  effects  on  bridge  trusses. 

(4)  Moving  load.  —  The  most  dangerous  moving  load  for  a 
highway  bridge   is   usually  a   crowd   of   people   or    a   drove   of 
animals.     This  is  commonly  taken  as  a  uniformly  distributed 
load,  which  may  cover  the  whole  bridge  or  any  portion  of  it. 
Its  value  is  variously  taken  at  from  60  Ibs.  to  100  Ibs.  per  square 
foot  of  area  of  floor,  depending  upon  the  span  and  upon  local 
conditions. 

It  may  be  that  in  certain  cases  the  greatest  stresses  will 
result  from  the  passage  of  heavy  pieces  of  machinery  over  the 
bridge,  as,  for  example,  a  steam  road  roller.  This  should  of 
course  be  considered  in  the  design. 

For  a  complete  discussion  of  loads  on  highway  bridges,  the 
student  is  referred  to  WaddeH'o  "  Highway  Bridges." 

136.  Loads  on  Railway  Bridges.  —  (i)  Permanent  load. — 
The  permanent  load  on  a  railway  bridge  includes  (a)  the  weight 
of  the  track  system,  which  is  known  or  may  be  determined  at 
the  outset ;  (b)  the  weight  of  longitudinal  stringers  and  cross- 


II4  GRAPHIC   STATICS. 

beams,  which  can  be  determined  before  the  trusses  are  designed  ; 
and  (c)  the  weights  of  trusses.  The  weight  of  the  track  system 
may  be  taken  at  400  Ibs.  per  linear  foot  for  a  single  track. 
(See  Burr's  "  Stresses  in  Bridge  and  Roof  Trusses  "  ;  Cooper's 
"Specifications  for  Iron  and  Steel  Railroad  Bridges";  Merri- 
man's  "  Roofs  and  Bridges.")  The  total  weight  of  track  system 
and  supporting  beams  and  stringers  varies  from  450  Ibs.  to 
600  Ibs.  per  linear  foot.  (Merriman.)  For  spans  less  than  100 
feet,  Merriman  gives  the  following  formulas,  in  which  w  is  the 
total  dead  load  of  the  bridge  in  pounds  per  linear  foot,  and  /  is 
the  span  in  feet  : 

For  single  track,   w=    560  +    5.6  /. 

For  double  track,  w=  1070+  10.7  /. 
See  also  Art.  8  of  Burr's  work  above  cited. 

(2)  Snow  and  wind,  —  Railway  bridges    usually  offer  little 
opportunity  for  the  accumulation  of  snow.     Wind  pressure  is, 
however,  an  important  factor.     Besides  the  pressure  upon  the 
bridge  itself,  the  pressure  upon  trains  crossing  the  bridge  must 
be  considered.     The  latter  is  a  moving  load  and  may  be  dealt 
with  in  the  same  way  as  other  moving  loads.     Its  amount  may 
be  computed  from  the  area  of  the  exposed  surface  of  the  train 
and  the  known   (or  assumed)   greatest   pressure   due  to  wind 
striking  a  vertical  surface  (Art.  85). 

For  further  discussion  of  wind  pressure,  the  student  is  referred 
to  the  works  already  cited.  The  graphic  methods  of  deter- 
mining stresses  due  to  wind  will  be  evident  when  the  methods 
for  vertical  loads  given  in  the  following  articles  are  understood. 

(3)  Moving  loads.  —  The  moving  load  to  be  supported  by  a 
bridge  consists  of  the  weights  of  trains.     Such  a  load  is  applied 
to  the  track  at  a  series  of  points,  namely,  the  points  of  contact 
of  the  wheels.     But  the  load  is  applied  to  the  trusses   only  at 
the  points  at  which  the  floor  beams  are  supported.     Hence  the 
actual  distribution  of  loads  upon  the  truss  is  somewhat  com- 
plex.    It  was   formerly  common    to   substitute  for   the  actual 


TRUSS  REGARDED  AS  A  BEAM.          H5 

load  a  uniformly  distributed  load,  thus  simplifying  the  problem 
of  determining  stresses.  It  is  now  more  usual  to  consider  the 
actual  distribution  of  loads  for  some  standard  type  of  locomo- 
tive used  by  the  railroad  concerned,  or  specified  by  its  engi- 
neers. For  examples  of  such  distributions  the  student  is 
referred  to  Cooper's  "  Specifications "  already  cited ;  also  to 
Fig.  47,  and  to  the  following  portions  of  this  chapter. 

137.  Through  and  Deck  Bridges.  — A  bridge  is  called  through 
or  deck  according  as  the  floor  system  is  supported  at  points  of 
the  lower  or  of  the  upper  chord.     In  the  former  case,  if  the 
trusses  are  too  low  to  require  lateral  bracing  above,  they  are 
called  pony  trusses. 

The  weight  of  the  truss  itself  is  to  be  divided  between  the 
upper  and  lower  joints.  But  the  weight  of  the  floor  system 
and  of  the  supporting  beams  and  stringers  comes  wholly  at  the 
lower  joints  of  a  through  bridge,  or  at  the  upper  joints  of  a 
deck  bridge.  The  moving  load  is,  of  course,  applied  at  the 
same  joints  at  which  the  floor  system  is  supported. 

If  the  floor  system  is  supported  directly  upon  the  upper 
chord,  as  is  sometimes  the  case,  the  moving  load  and  part  of 
the  dead  load  produce  bending  in  the  chord  members  ;  other- 
wise the  design  is  unaffected  by  this  construction. 

§  2.    Truss  Regarded  as  a  Beam. 

138.  Classification  of  Trusses.  —  Since  a  bridge  truss  acts  as 
a  practically  rigid  body  resting  on  supports  at  the  ends  or  other 
points  and  sustaining  vertical  loads,  it  may  be  regarded  as  a 
beam,  and  trusses  may  be  classified  in  the  same  way  as  beams 
(Art.   114).     The  only  class  to  be  here  considered   is  that  of 
simple  trusses,  or  such  as  may  be  regarded  as  rigid  bodies  for 
the  purpose  of  determining  the  reactions. 

Cantilever  trusses  and  continuous  trusses  are  defined  like  the 
corresponding  classes  of  beams  (Art.  114).  The  most  impor- 
tant case  is  that  of  a  truss  simply  supported  at  the  ends. 


GRAPHIC    STATICS. 


iM 


\ 


Fig.  40 


139.  External  Shear  for  a  Truss.  —  If  a  truss  be  regarded  as 
a  beam,  the  external  shear,  resisting  shear,  and  internal  shear- 
ing stress  at  any  section  may  be  defined  just  as  in  Art.  115. 
In  some  forms  of  truss  a  knowledge  of  the  external  shear  at 
any  section  makes  it  possible  to  compute  readily  the  stresses  in 
certain  truss  members.     Thus,  in  the  portion  of  a  truss  repre- 
sented in   Fig.  49,  let  the  section 
MN  cut  three  members,  of  which 
two  are  horizontal.     (The  member 
x'y'   is    disregarded.)      Since    each 
member  can    exert   forces   only    in 
the  direction  of  its  length,  the  ex- 
ternal   shear   in   the    section    MN 
must  be  wholly  resisted  by  the  di- 
agonal xy ;  and  the  internal  force 
in  xy  must  be  such  that  its  resolved 

part  in  the  vertical  direction  is  equal  in  magnitude  to  the  exter- 
nal shear  in  the  section.  Let  the  external  shear  V  be  repre- 
sented by  YZ  (Fig.  49)  ;  draw  YX  parallel  to  yx  and  ZX 
horizontal;  then  JfFwill  represent  the  stress  in  xy.  If  V  is 
positive  (Art.  115),  the  stress  in  xy  is  a  tension.  If  Fis  nega- 
tive, the  stress  is  a  compression.  If  the  member  xy  were 
replaced  by  one  sloping  the  other  way  from  the  vertical,  these 
statements  as  to  kind  of  stress  would  be  reversed. 

If  no  two  of  the  three  members  cut  by  any  section  are  par- 
allel, the  stresses  cannot  be  computed  so  simply,  since  all  may 
contribute  components  of  force  to  resist  the  external  shear. 

140.  Bending   Moment  for   a   Truss.  —  In   many   cases   the 
stresses  in  the  truss  members  can  be  found  from  the  values  of 
the  bending  moment  at  different  sections. 

Thus,  in  Fig.  49,  let  a  section  MN  be  taken  cutting  three 
members  as  shown,  and  let  the  origin  of  moments  be  taken  at 
the  point  of  intersection  of  two  of  them  (as  bx  and  xy)  ;  then 
since  the  moments  of  the  internal  forces  in  these  two  will  be 


PARALLEL   CHORDS  —  CONCENTRATED    LOADS.          117 

zero,  the  resisting  moment  is  equal  to  the  moment  of  the  inter- 
nal force  in  the  third  member  ay.  The  arm  of  this  latter  force 
is  the  perpendicular  distance  of  its  action  line  from  the  origin. 
Call  it  k,  and  let/  represent  the  force  itself,  and  J/the  bending 
moment.  Then,  numerically, 


If  M  is  positive  (in  accordance  with  the  convention  of  Art. 
47),  p  must  act  from  right  to  left,  hence  the  stress  in  ay  is  a 
compression.  If  M  is  negative,  the  stress  is  a  tension.  (It 
must  be  remembered  that  the  forces  whose  moments  make  up 
the  bending  moment  M  act  upon  the  portion  of  the  truss  to 
the  left  of  the  section.) 

This  method  applies  equally  well  to  the  case  in  which  no  two 
of  the  members  cut  are  parallel.  It  is  an  application  of  a  prin- 
ciple of  much  importance  in  the  determination  of  stresses  in 
truss  members. 


§  3.    Truss  witk   Parallel  CJiords  sustaining  Concentrated 

Loads. 

141.  General  Method.  — It  was  shown  in  Art.  139  that  when 
both  chords  of  the  truss  are  horizontal,  so  that  a  vertical  sec- 
tion through  the  truss  will  cut  but  one  inclined  member,  the 
stress  in  such  an  inclined  member  may  be  easily  found  from  the 
external  shear  in  the  section.  Hence  for  such  a  truss,  we  may 
determine  the  greatest  stresses  in  the  web  members  by  con- 
structing the  curve  of  maximum  shear  for  the  truss,  in  a  man- 
ner similar  to  that  employed  in  Art.  127  for  a  beam. 

Again,  the  curve  of  maximum  bending  moments  will  usually 
enable  us  to  determine  the  greatest  chord-stresses  (Art.  140). 

The  construction  of  these  curves  for  the  case  of  a  truss  is 
not  quite  so  simple  as  in  case  of  a  beam,  for  reasons  to  be 
explained  in  the  following  article. 


H8  GRAPHIC   STATICS. 

142.  Distribution  of  Loads  on  Truss.  —  If  the  curves  of  maxi- 
mum shear  and  moment  for  a  given  series  of  concentrated  loads 
be  constructed  as  in  Arts.  127  and  130,  they  will  not  represent 
correctly  the  maximum  shear  and  moment  at  all  points  of  the 
truss.     The  reason  for  this  will  be  seen  by  considering  the  way 
in  which  the  loads  are  actually  applied  to  the  truss.     The  road- 
way of  the  bridge  is  usually  supported  by  longitudinal  beams, 
which  themselves  rest  upon  cross-beams  or  girders,  the  latter 
being  supported  at  the  joints  of  the  truss.     Hence,  however 
the  loads  may  be  distributed  upon  the  roadway,  they  are  borne 
by  the  truss  only  at  the  joints.     But  in  the   method  used  in 
Arts.   127  and    130  for  drawing  curves  of  shear  and  bending 
moment,  it  was  assumed  that  every  load  rests  directly  upon  the 
beam,  at  a  point  immediately  under  the  load. 

In  the  case  of  the  truss,  since  loads  are  applied  only  at  the 
joints,  the  shear  at  any  instant  is  the  same  throughout  the 
length  of  a  panel.  What  is  needed,  therefore,  is  the  greatest 
possible  shear  for  each  panel  of  the  truss.  For  simplicity,  we 
shall  consider  first  the  shear  due  to  dead  loads,  and  then  that 
due  to  the  live  loads,  taking  a  numerical  example. 

143.  Problem  —  Numerical    Data.  —  Assume    a    " through" 
truss  of  the  form  shown  in  skeleton  in  Fig.  50  (PL  IV)  ;  let  the 
span  be  96  ft.,  the  truss  being  divided  into  6  panels  of  16  ft. 
each.     The  depth  is  taken  at  15  ft.     Assume  the  dead  load  as 
1 100  Ibs.  per  linear  foot  for  the  whole  bridge,  — one-half  being 
borne  by  each  truss,  — of  which  one-third  is  the  weight  of  the 
floor  system  and  track,  and  is  therefore  borne  by  the  lower 
chord,  and  the  remaining  two-thirds  is  assumed  to  be  divided 
equally  between  the  upper  and  lower  chords.     We  have  then 
for  the  dead  loads  : 

Lower  panel  load  =  5 866  Ibs.,  say  6000  Ibs. 
Upper  panel  load  =  293 3  Ibs.,  say  3000  Ibs. 

For  the  moving  load  we  will  assume  a  train  drawn  by  two 
locomotives  with  weights  distributed  as  shown  in  Fig.  50.  In 


PARALLEL   CHORDS —CONCENTRATED    LOADS.          119 

computing  stresses  the  train  will  be  supposed  to  come  on  the 
bridge  from  the  right. 

144.  Shear  Curve  for  Fixed  Loads.  —  The  curve  of  shears  for 
the  dead  load  may  be  constructed  as  follows :  In  Fig.  50  (PI.  IV), 
the   line  XSYS  represents  the  axis  of  abscissas  for  the  shear 
curve.     The  reaction  at  the  left  support,  minus  the  dead  load 
borne  at  that  point,  gives  24000  Ibs.  as  the  positive   shear  for 
every  section  between  the  support  and  the  first  upper  panel 
joint.     This  is  laid  off  downward  from  XSYS.     Then  passing  to 
the  right,  as   each  load  is  passed,  its  amount   must    be    sub- 
tracted from  the  shear.     The  result  is   the  broken    line  X'Y' 
shown  in  the  figure. 

145.  Moment   Curve  for   Fixed  Loads.  —  To    construct  the 
bending  moment  curve  for  dead  loads,  we  have  only  to  draw  a 
funicular  polygon  for  these  loads  and  the  reactions  due  to  them ; 
the  vertical  ordinate  at  any  section  will  represent  the  bending 
moment  at    that    section.     The    actual   value   of    the    bending 
moment  is  the  product  of  the  ordinate  into  the  pole  distance  ; 
the  former  being  measured  in  linear  units,  the  latter  in  force 
units.     The  curve  is  shown  in  Fig.  50  (PL  IV),  the  ordinates 
being  laid  off  upward  from  the  line  Xm  Ym.     The  force  polygon 
is  shown  at    IV,  the  pole  being   O,  taken  opposite  the  middle 
point  of  the  load-line  in  order  that  the  closing  side  of  the  funic- 
ular polygon  may  be  horizontal,  and  can  therefore  be  made  to 
coincide  with  XmYm.     The  pole  distance  represents  120,000  Ibs. 

146.  Actual  Shear  for   Given   Position   of   Moving  Loads.  - 

The  shear  in  any  panel  is  readily  found  when  the  position  of 
live  loads  is  known.  We  have  only  to  apply  the  definition  of 
external  shear  (Art.  115)  as  the  sum  of  all  vertical  forces  acting 
on  the  portion  of  the  truss  to  the  left  of  the  section  con- 
sidered. That  is,  we  must  find  the  reaction  at  the  left  abut- 
ment and  subtract  from  it  the  sum  of  the  loads  between  the 
left  abutment  and  the  section.  It  must  be  noticed  that  any 


120  GRAPHIC    STATICS. 

load  on  the  panel  considered  is  divided  between  the  two  adja- 
cent joints ;  the  part  supported  at  the  left  end  of  the  panel  is 
to  be  subtracted  in  computing  the  shear.  The  graphic  con- 
struction is  explained  in  the  following  articles. 

[NOTE.  —  It  will  be  noticed  that  the  actual  distribution  of  any  load  among  the 
different  panel  points  is  not  strictly  determinate  by  methods  heretofore  treated,  if  the 
longitudinal  beams  supporting  the  floor  system  are  continuous  (supported  at  more 
than  two  points.  See  Art.  114).  We  shall  assume  the  portion  of  such  a  beam 
between  two  supports  to  act  as  a  simple  beam.  The  results  of  this  assumption  are 
probably  as  reliable  as  could  be  obtained  by  a  more  elaborate  discussion.] 

147.  Position  of  Moving  Load  for  Maximum  Shear  in  Any 
Panel. — The  position  of  the  moving  load  which  will  give  the 
maximum  shear  in  any  panel  may  be  determined  graphically  by 
repeated  trials.  But  it  is  possible  to  deduce  a  simple  rule  which 
will  shorten  the  labor  very  materially. 

It  must  first  be  noticed  that  the  general  principles  stated 
in  Art.  126,  in  treating  of  beams,  are  equally  applicable  here. 
We  know,  therefore,  in  a  general  way,  that  in  order  to  get  the 
greatest  positive  shear  in  a  given  panel,  the  portion  of  the  truss 
to  the  right  should  be  loaded  as  completely  as  possible,  while 
the  part  to  the  left  should  be  free  from  loads.  This  rule  is  not, 
however,  absolute.  In  general  the  foremost  loads  will  be  upon 
the  panel  in  question,  or  possibly  to  the  left  of  it,  in  the  posi- 
tion of  greatest  shear.  We  shall  now  prove  the  following 

Proposition.  —  When  the  shear  in  any  panel  has  its  greatest 
value,  the  load  upon  that  panel  is  to  the  total  load  on  the  truss 
as  the  length  of  the  panel  is  to  the  length  of  the  truss. 

Let  JfF(Fig.  51)  represent  the  length  of  the  truss  (=/),  and 
CD  (  =  /')  the  length  of  the  panel.  Let  W  =  total  load  on 
truss,  and  W  =  total  load  on  the  panel.  Consider  the  effect  of 
moving  all  the  loads  to  the  left.  Evidently  the  reaction  at  X 
is  increased  and  the  load  at  C  is  increased.  The  former 
increases  the  shear  in  the  panel,  while  the  latter  decreases  it. 
These  are  the  only  changes  which  are  caused  in  the  shear  by 
the  movement  of  the  load.  Now  if  the  position  of  loads  is 


PARALLEL   CHORDS  —  CONCENTRATED    LOADS.          I2i 

such  that  the  shear  is  a  maximum,  the  two  effects  mentioned 
must  just  neutralize  each  other  for  a  very  small  displacement  of 
the  loads.  (Otherwise  a  small  displacement  in  one  direction  or 
the  other  would  increase  the  shear.) 


A' 


Let  x  =  distance  of  center  of  gravity  of  W  from  Fand  ,r'  = 
distance  of  center  of  gravity  of  IV  from  D.  Let  R  =  reaction 
at  X  and  P  =  load  at  C  due  to  IV.  Then 

R=  W-\  P=  IV^- 

If  the  whole  load  moves  any  distance,  x  and  x'  both   increase 
by  the  same  amount.     For  an  infinitesimal  displacement, 

JP      W,       rD     IV  ,        W, 
dR  = —  dx\  dP  =  —dx=  —  dx. 

Now  the  increment  of  the  shear  is  dR  —  dP  or 

(W     IV\    , 

IT— J  ** 

Since  this  must  be  zero  (as  stated  above),  we  have 
IV=W  or    W_  =  P_ 

r  ~~~~  /'      w    i* 

for  the  position  of  maximum  shear.     The  proposition  is  there- 
fore proved. 

The  condition  may  be  written    W=—W*.       If    the  truss    is 

/ 
divided  into  n  equal  panels,  then  jf  =  n,  and  W=n  IV. 

This  principle  is  so  simple  that  it  is  easier  of  application  than 
the  graphic  method  of  repeated  trials. 


122  GRAPHIC   STATICS. 

[NOTE.  —  In  the  above  discussion  it  is  assumed  that  IV  and  IV  both  remain  con- 
stant during  the  movement  dx;  that  is,  that  no  load  enters  or  leaves  the  panel  or  the 
whole  truss.  Hence  the  reasoning  is  not  rigorous  for  a  case  in  which  the  condition 
W=  -  IV  requires  a  load  to  be  at  Cor  D,  or  at  X  or  F.  The  remark  made  in  the 
note  in  Art.  129  applies  here.  The  condition  W  =  —  IV  may  always  be  applied  pro- 
vided a  load  at  any  joint  may  be  treated  as  if  divided  between  the  adjacent  panels  in 
any  desired  ratio.] 

148.  Construction  of  Curve  of  Maximum  Shear.  —  In  Fig.  50 
(PI.  IV)  are  shown  the  lines  of  action  of  the  series  of  moving 
loads  to  be  carried  by  the  truss.  The  force  polygon  or  load 
line  is  drawn  for  the  whole  series  of  loads,  being  the  line 
ABC.  .  .  .  For  the  purpose  of  drawing  the  funicular  polygon, 
the  uniformly  distributed  load  is  divided  into  parts  of  6000  Ibs., 
each  assumed  to  act  at  its  center  of  gravity.  With  a  pole  O 
(the  pole  distance  being  taken  so  as  to  represent  120000  Ibs.) 
the  funicular  polygon  is  drawn  for  the  series  of  loads.  Let  us 
now  consider  the  greatest  shear  in  any  panel  of  the  truss,  as  in 
KL  (Fig.  50). 

Applying  the  principle  already  deduced,  it  is  easily  shown 
that  the  greatest  shear  occurs  when  the  load  CD  is  at  the  point 
L.  For,  suppose  CD  is  just  at  the  left  of  L  ;  the  total  load  on 
the  panel  is  then  38000  Ibs.,  while  the  total  load  on  the  truss 
is  172000  Ibs.  Now  the  ratio  of  the  whole  span  to  the  panel 

length  r~,  Art.    147  j  is  6 ;  and  since  38000  Ibs.  is  greater  than 

one-sixth  of  172000  Ibs.,  the  load  on  the  panel  is  too  great  to 
satisfy  the  required  condition.  Again,  suppose  CD  is  just  at 
the  right  of  L  ;  the  total  load  on  the  panel  is  23000  Ibs.,  which 
is  less  than  one-sixth  of  172000;  hence  the  load  on  the  panel  is 
too  small  to  satisfy  the  condition  for  maximum  shear.  We  con- 
clude, therefore,  that  the  shear  is  a  maximum  when  the  load 
CD  is  just  at  the  point  L.  In  Fig.  50,  S2S2  represents  thet 
beam  with  load  CD  at  L.  Drawing  vertical  lines  through  the 
ends  of  S2S2,  the  points  in  which  these  intersect  the  strings  oa 
and  of  determine  two  points  of  the  closing  string  (0/2)  of  the 
funicular  polygon  for  the  loads  and  reactions  now  acting  on  the 


PARALLEL  CHORDS —  CONCENTRATED  LOADS. 


123 


truss.  The  ray  (9/2  is  drawn  parallel  to  this  closing  line,  inter- 
secting the  load  line  in  _/2.  This  determines  JZA  as  the  left 
reaction.  To  get  the  shear  in  the  panel,  we  must  subtract  from 
this  reaction  the  portion  of  the  two  loads  AB,  BC,  which  comes 
upon  the  truss  at  K.  This  portion  is  found  by  treating  KL  as 
a  simple  beam  and  determining  the  reactions  at  K  and  L  due  to 
loads  AB  and  BC.  Draw  vertical  lines  through  the  points  K 
and  L  of  the  truss  (in  the  position  S2S2)  ',  the  points  in  which 
these  verticals  intersect  the  strings  oa  and  oc  are  two  points 
of  the  closing  side  of  the  funicular  polygon  for  the  loads 
and  reactions  on  the  beam  KL.  Parallel  to  this  closing  side 
draw  a  ray  OJJ.  This  line  divides  the  load  line  AC  into  two 
segments  representing  the  portions  of  AB  and  BC  supported 
on  the  truss  at  K  and  L  respectively.  Hence  the  shear  in 
the  panel  is  the  reaction  J2A  minus  the  load  represented  by 
the  line  AJ«. 

Taking  XsYg  (Fig.  50)  as  the  axis  of  abscissas  for  the  curve 
of  shears,  the  portion  of  the  curve  of  maximum  shear  corre- 
sponding to  the  panel  KL  is  a  straight  line  parallel  to  Xt  Ys,  at  a 
distance  from  it  equal  toJ-2J-2. 

The  curve  of  greatest  shear  for  the  whole  truss  is  shown  in 
the  figure,  being  the  broken  line  X"  Yst  but  the  construction  is 
given  only  for  the  panel  KL.  For  other  panels  the  method 
will  be  exactly  similar. 

149.    Position   of    Load   for    Maximum   Bending    Moment.— 

In  computing  stresses  in  the  chord  members,  the  greatest 
bending  moment  is  needed  for  each  vertical  section  containing 
a  chord  joint.  In  considering  the  position  of  moving  loads 
which  will  give  the  greatest  bending  moment  at  any  section, 
the  sections  through  joints  carrying  live  loads  must  be  treated 
separately  from  those  through  joints  free  from  loads. 

Let  Fig.  51  represent  a  through  truss  (Art.  137),  and  con- 
sider first  the  bending  moment  at  a  section  containing  one 
of  the  lower  joints,  as  C.  It  is  readily  seen  that  for  such  a 


124 


GRAPHIC   STATICS. 


section  the  bending  moment  due  to  a  load  anywhere  on  the 
truss  is  exactly  the  same  as  if  the  truss  were  a  beam  upon 
which  the  load  was  supported  directly.  For,  consider  a  load,  on 
any  panel  as  £>C.  Though  the  load  is  actually  divided  between 
B  and  C,  the  reaction  it  causes  at  X  is  the  same  as  if  the  truss 
were  a  beam  supporting  the  load  directly.  Also,  since  the  load 
is  the  resultant  of  the  two  parts  at  B  and  C,  the  sum  of  the 
moments  of  these  two  parts  about  any  origin  is  equal  to 
the  moment  of  the  load  itself.  Now,  in  computing  bending 
moment  caused  by  the  given  load  at  any  section,  we  take  the 
moment  of  the  left  reaction  minus  the  part  of  the  load  which 
is  carried  at  the  left  of  the  section.  If  the  section  is  anywhere 
except  between  B  and  C,  the  result  is  the  same  whether  we  use 
the  given  load  in  its  actual  position,  or  its  two  components  at 
B  and  C. 

For  finding  the  greatest  bending  moments  at  the  sections  A, 
B,  C,  etc.,  we  may  therefore  apply  the  same  rule  as  in  the  case 
of  a  beam  (Art.  129).  That  is,  when  the  bending  moment  for 
such  a  section  is  a  maximum,  the  loads  on  the  two  segments  of 
the  truss  are  proportioned  to  the  lengths  of  the  segments. 

Next  consider  a  section  somewhere  between  two  loaded 
joints,  as  at  C  (Fig.  51).  The  principles  above  stated  hold  for 
such  a  section  for  loads  anywhere  except  on  the  panel  BC.  In 
case  of  a  load  on  this  panel,  only  a  part  is  borne  at  the  left  of 
C ;  and,  in  computing  bending  moment,  the  part  coming  at  C 
is  to  be  omitted.  Hence  the  above  reasoning  is  not  strictly 
applicable  to  this  case.  The  following  modification  of  the 
principle  above  deduced  may  however  be  shown  to  hold  for  any 
section  of  the  truss. 

Let  the  given  section  divide  a  panel  BC  into  segments  BC" 
and  C" C.  Then,  if  all  loads  on  BC  are  treated  as  if  divided 
between  B  and  C  in  the  ratio  of  BC"  to  C"C,  the  general  rule 
applicable  to  a  beam  may  be  used  in  assigning  the  position  of 
loads  for  maximum  bending  moment. 

[The  proof  of  this  proposition  will  be  omitted.     The  mode 


PARALLEL   CHORDS  — CONCENTRATED    LOADS.          125 

of  reasoning  to  be  applied  is  similar  to  that  used  in  Art.  147. 
The  student  should  attempt  the  proof  for  himself.] 

In  general,  the  error  will  be  small  if  the  bending  moment 
curve  for  the  case  of  a  beam  is  used  in  computing  bending 
moments  at  all  panel  joints  of  the  truss. 

150.  Curve  of  Maximum  Bending  Moments.  —  The  greatest 
bending  moment  at  any  section  can  be  easily  found  as  soon 
as  the  position  of  moving  loads  producing  it  is  known.  The 
method  will  be  explained  with  reference  to  the  truss  shown  in 
Fig.  50  (PI.  IV). 

By  Art.  149,  the  greatest  bending  moment  for  each  of  the 
points  K,  L,  M,  N,  P,  is  exactly  the  same  as  if  the  truss  were 
a  beam  upon  which  the  loads  were  supported  directly.  It  is 
unnecessary  to  show  the  construction  for  these  points,  it  being 
identical  with  that  of  Art.  130.  We  therefore  consider  a  sec- 
tion through  an  upper  joint  as  T.  First  the  position  of  the 
moving  load  must  be  assigned. 

By  Art.  149,  we  may  apply  the  general  principle  that  the 
loads  in  the  two  segments  of  the  truss  should  be  in  the  same 
ratio  as  their  lengths,  provided  all  loads  on  the  panel  MN  are 
treated  as  if  equally  divided  between  J/and  N  (since  the  sec- 
tion considered  divides  MN  into  equal  parts).  Again,  since 
the  segment  of  the  truss  to  the  left  of  the  section  is  seven- 
twelfths  of  the  whole  span,  the  load  on  the  left  segment  should 
be  seven-twelfths  the  total  load  on  the  truss. 

It  is  found  by  trial  that  the  condition  for  maximum  bending 
moment  at  T  is  satisfied  when  the  line  of  action  b'c  is  at  M. 
For,  the  total  load  on  the  truss,  when  the  load  is  near  this  posi- 
tion, is  182000  Ibs.  Now  suppose  the  load  B*O  is  just  at  the 
left  of  M.  The  three  loads,  CD\  D'E\  E'F',  must  be  regarded 
as  equally  divided  between  M  and  N ;  this  gives  for  the  whole 
load  to  the  left  of  T  111500  Ibs.,  which  is  greater  than  seven- 
twelfths  of  182000  Ibs.  But  if  B'C  is  just  at  the  right  of  M, 
half  of  it  must  be  regarded  as  borne  at  N,  and  the  whole  load 


126  GRAPHIC   STATICS. 

on  the  truss  to  the  left  of  T  is  104000  Ibs.,  which  is  less  than 
seven-twelfths  of  182000  Ibs. 

Now  draw  Mt  Mt  to  represent  the  beam  in  the  position  for 
maximum  bending  moment  at  T\  that  is,  with  the  load  B'C  at 
M.  From  its  extremities  draw  vertical  lines  intersecting  the 
strings  od  and  0c",  and  through  the  points  thus  determined 
draw  ot\  the  closing  side  of  the  funicular  polygon  for  all  the 
loads  and  reactions  acting  on  the  truss. 

In  determining  the  bending  moment  by  the  method  of  Art. 
130,  we  must  consider  the  loads  on  the  panel  MN  as  replaced 
by  their  components  at  M  and  N  respectively.  Draw  vertical 
lines  through  ^/and  A7"  intersecting  the  strings  oc'  and  of,  and 
join  the  points  thus  determined.  The  line  thus  drawn  should 
replace  the  corresponding  portion  of  the  funicular  polygon 
before  drawn  (i.e.,  the  strings  oc\  od',  oe\  and  a  part  of] ; 
because  it  is,  in  fact,  a  line  of  the  funicular  polygon  which  will 
be  obtained  if  CD',  D'E',  and  E'F'  be  replaced  by  their  com- 
ponents which  are  actually  applied  to  the  truss  at  M  and  N. 
We  now  draw  through  T  a  vertical  line  intersecting  ot'  and  the 
corrected  string  just  determined.  The  intercept  thus  deter- 
mined, multiplied  by  the  pole  distance,  gives  the  bending 
moment  at  T. 

From  Xm  Ym  as  a  line  of  reference  draw  downward  an  ordi- 
nate  at  the  point  T  equal  to  the  intercept  thus  found ;  this 
gives  a  point  of  the  curve  of  maximum  bending  moments. 

Other  points  of  the  curve  may  be  found  in  the  same  way. 
The  whole  curve  is  shown  in  the  figure,  but  the  construction  is 
not  given  except  for  the  section  T. 

[NOTE. — The  condition  for  maximum  moment  at  a  section  may  sometimes  be 
satisfied  for  more  than  one  position  of  the  moving  load.  This  is  the  case  for  the 
section  at  T\  it  will  be  found  by  trial  that  the  condition  is  satisfied  when  the  load 
O'D"  is  just  at  the  right  end  of  the  truss.  The  bending  moment  for  this  position 
is,  however,  found  to  be  slightly  less  than  for  the  position  above  taken.  In  most 
cases,  the  position  for  greatest  moment  can  be  very  nearly  predicted  by  remembering 
that  the  truss  should  be  loaded  as  completely  as  possible,  and  that  the  heavy  loads 
should  be  as  near  the  section  as  possible.  These  principles  will  also  serve  as  a  guide 
in  distinguishing  whether  the  moment  is  a  maximum  or  a  minimum  when  the  condi- 
tion is  satisfied.] 


PARALLEL   CHORDS  —  CONCENTRATED    LOADS.          127 

It  should  be  noticed  that  the  value  found  for  the  bending 
moment  at  any  section  is  a  possible  value  for  another  section 
equally  distant  from  the  middle  point  of  the  truss,  since  the 
train  may  cross  the  bridge  from  the  left  instead  of  from  the 
right.  Hence  for  two  points,  as  L  and  N,  equally  distant  from 
the  middle  of  the  truss,  the  greater  of  the  two  moments  above 
determined  must  be  used  for  both. 

151.  Shear  and  Moment  Curves  for  Combined  Fixed  and  Mov- 
ing Loads.  —  In  Fig.  50  (PI.  IV)  the  line  X  Ys  has  been  taken 
as  the  axis  of  abscissas  for  the  curves  of  shear.     Positive  shear 
has  been  laid  off    downward  for  fixed  loads  and  upward  for 
moving  loads.     Hence  an  ordinate  at  any  point  drawn  perpen- 
dicular to  Xs  Ys,  and  limited  by  the  two  curves  (or  rather  broken 
lines),  represents  the  greatest  positive  shear  due  to  combined 
fixed  and  moving  loads.     The  shear  curves  for  fixed  and  mov- 
ing loads  intersect  at  a  point  directly  opposite  the   joint  N. 
Hence  for  sections   to  the  right   of   this   point,  the    shear  is 
always  negative. 

Similarly,  Xm  Ym  is  the  axis  of  abscissas  for  the  two  moment 
curves.  The  bending  moment  is  always  negative ;  in  case  of 
the  dead  load  curve,  it  is  laid  off  upward  from  XmYm>  while  for 
the  live  load  curve  it  is  laid  off  downward.  Hence  an  ordinate 
between  the  live  and  dead  load  curves  at  any  point  of  Xm  Ym 
represents  the  greatest  possible  bending  moment  at  the  cor- 
responding section  of  the  truss. 

We  are  now  prepared  to  determine  maximum  stresses  in  the 
truss  members. 

152.  Maximum   Stresses   in  Web   Members.  —  The  greatest 
stress    in  any  web   member  is  readily  found  from  the   shear 
curve.     By   Art.    139,   the    stress    has    such    a   value   that    its 
resolved  part  in  the  vertical  direction  is  equal  to  the  shear  in  a 
section  through  the  member.     For  example,  for  the  member 
14-15  we  draw  a  line  in  the  shear  diagram  parallel  to  14-15  and 
limited  by  the  two  lines  which  determine  the  greatest  shear 


I2g  GRAPHIC   STATICS. 

sustained  by  the  member.  This  gives  14'-! 5'  as  the  stress  in 
the  given  member.  In  the  same  way,  I2r-i3',  1^-14',  etc., 
represent  stresses  in  the  corresponding  truss  members. 

As  to  the  sign  of  the  stress  in  any  member,  we  know  that 
for  a  positive  shear  any  diagonal  member  has  to  resist  a  ten- 
dency of  the  portion  of  the  truss  to  the  left  to  move  upward ; 
hence  the  stress  is  a  tension  for  every  member  sloping  down- 
ward from  left  to  right  (as  13-14),  and  a  compression  for  a 
member  sloping  downward  from  right  to  left  (as  14-15).  The 
reverse  will  be  true  if  the  shear  is  negative.  We  may,  how- 
ever, neglect  the  negative  shears  given  by  the  curve  of  Fig.  50, 
because  they  are  the  minimum  negative  shears.  (This  is  evi- 
dent because  they  were  determined  as  maximum  positive 
shears;  and  since  they  are  found  to  be  really  negative,  they 
are  the  least  negative  shears  that  can  occur  at  those  sections.) 
The  maximum  negative  shear  for  this  portion  of  the  truss  will 
be  found  when  the  train  crosses  the  bridge  from  left  to  right ; 
and  its  value  for  any  section  is  the  same  as  that  of  the  greatest 
positive  shear  for  the  section  equally  distant  from  the  middle  of 
the  truss  but  on  the  other  side. 

For  example,  since  we  have  found  a  negative  shear  in  sec- 
tions through  the  member  21-22,  that  member  sustains  a  com- 
pression which  is  represented  in  the  figure  by  21 '-2 2'.  But  if 
the  train  is  headed  to  the  right,  the  member  21-22  may  sustain 
a  much  greater  compression  —  equal,  in  fact,  to  that  already 
found  for  14-15,  and  represented  by  14'-!  5'.  The  latter  is  there- 
fore the  greatest  compression  sustained  by  21-22. 

It  is  evident,  therefore,  that  it  is  necessary  to  determine 
stresses  from  Fig.  50  for  those  members  only  which  sustain 
positive  shear;  remembering  that  the  stress  thus  found  in  any 
member  is  a  possible  value  of  the  stress  in  the  corresponding 
member  on  the  other  side  of  the  center  of  the  truss. 

If  both  maximum  and  minimum  stresses  are  desired,  it  must 
be  remembered  that  the  dead-load  stresses  may  occur  alone. 

It  will  be  seen  that  a  reversal  of  stress  may  occur  in  the 


PARALLEL   CHORDS  —  CONCENTRATED    LOADS.          129 

members  16-17,  17-18,  18-19,  19-20;  while  in  the  remaining 
members  no  reversal  is  possible. 

153.  Maximum  Chord  Stresses.  —  From  the  maximum  bend- 
ing moments  represented  in  Fig.  50,  we  may  readily  find  the 
greatest  stress  sustained  by  any  chord  member.  Consider, 
for  example,  the  member  15-2.  If  a  section  be  taken  cutting 
the  three  members  7-14,  14-15,  and  15-2,  and  the  origin  of 
moments  be  taken  at  R,  it  is  evident  that  the  sum  of  the 
moments  of  the  reaction  and  loads  to  the  left  of  R  is  equal  to 
the  moment  of  the  force  acting  in  the  member  15-2.  But  the 
former  is  the  bending  moment  in  the  section  through  R,  and 
its  greatest,  value  is  equal  to  the  .ordinate  between  the  two 
moment  curves  already  drawn,  multiplied  by  the  pole  distance. 
Hence  we  have  the  equation  : 

Ordinate  of  moment  curve  x  pole  distance  =  stress  in  chord 
member  x  depth  of  truss. 

That  is,  in  the  present  example,  the  stress  in  pounds  is  equal  to 
the  ordinate  multiplied  by  120000  and  divided  by  15.  The 
computation  may  be  made  graphically  as  follows  :  From  any 
point  Z  (Fig.  50)  draw  two  lines  making  a  convenient  angle 
with  each  other,  say  a  right  angle.  Lay  off  ZZn  equal  to  the 
pole  distance  on  the  force  scale  already  adopted,  and  ZZ1  equal 
to  the  depth  of  the  truss,  and  draw  Z' Z" .  Then  take  Zz  equal 
to  the  ordinate  of  the  moment  curve  and  draw  2-15  parallel  to 
Z* Z" ,  intersecting  ZZ11  in  the  point  marked  15.  The  distance 
Z-1%  represents  the  stress  in  15-2,  to  the  force  scale  already 
adopted. 

If  the  pole  distance  ZZ"  had  been  drawn  to  some  other  scale, 
Z-\$  would  represent  the  required  stress  to  the  same  scale. 

For  two  chord  members  symmetrically  situated  with  respect 
to  the  middle  of  the  truss,  the  maximum  stresses  have  the  same 
value,  which  is  to  be  computed  from  the  greater  of  the  two 
corresponding  values  of  the  bending  moment.  This  is  apparent, 
if  it  is  remembered  that  when  the  train  is  headed  to  the  right, 


130 


GRAPHIC   STATICS. 


the  two  members  interchange  their  conditions.  The  figure 
shows  all  the  chord  stresses.  Evidently,  all  upper  chord  mem- 
bers are  in  compression  and  all  lower  ones  in  tension. 

If  minimum  stresses  are  desired,  they  will  be  found  by  using 
the  dead-load  bending  moments  alone. 

The  computation  of  chord  stresses  will  be  facilitated  if  the 
number  of  force  units  in  the  pole  distance  is  taken  as  a  simple 
multiple  of  the  number  of  linear  units  in  the  depth  of  the 
truss.  For,  if  H  is  the  pole  distance,  //  .the  depth  of  truss,  y 
the  ordinate  of  the  moment  curve,  and  x  the  stress  in  the  chord 
member,  we  have 

Hy  =  hx ;  or  x  =  —-y. 

TT 

Now  if  —  =  «,  we  have  only  to  measure  y  in  linear  units,  multiply 
by  ;/,  and  the  result  is  x  in  force  units. 

§  4.     Parallel  Chords — Uniformly  Distributed  Moving-  Load. 

1 54.  Distribution  of  Load.  —  In  designing  highway  bridges, 
it  is  usual  to  assume  the  moving  load  to  be  uniformly  distributed 
along  the  bridge.  The  same  assumption  is  sometimes  made 
for  railroad  bridges.  This  simplifies  the  computation  of  stresses 
very  materially.  In  fact,  when  the  chords  are  parallel,  the 
algebraic  method  of  treating  such  cases  becomes  so  simple  that 
it  is  by  many  preferred  to  the  graphic  method.  It  will  be 
well,  however,  to  indicate  the  main  points  in  the  graphic  con- 
struction. 

Maximum  shear.  -. —  First  it  may  be  noticed  that  the  position 
of  moving  load  for  greatest  shear  in  any  panel  is  readily  deter- 
mined. The  conclusion  deduced  in  Art.  147  may  be  shown  to 
apply  to  this  case  ;  hence  we  have  the  principle  that  the  greatest 
shear  in  any  panel  occurs  when  the  portion  of  the  truss  to  the 
right  is  completely  loaded,  while  the  load  on  the  panel  is 
the  same  fraction  of  the  whole  load  that  the  panel  length 
is  of  the  whole  span.  Let  ArF(Fig.  51)  represent  a  truss  of 


PARALLEL   CHORDS  — DISTRIBUTED   LOAD.  ni 

seven  equal  panels ;  and  for  the  greatest  shear  in  a  panel  CD 
let  Z  be  the  foremost  point  of  the  moving  load.  Then  we  must 
have  ZD  equal  to  one-seventh  of  ZY\  or,  ZD  =  one-sixth  DY. 
A  similar  rule  applies  to  each  panel. 

Maximum  moment.  —  From  the  principles  deduced  in  Arts. 
129  and  149,  it  is  evident  that  the  greatest  bending  moment 
occurs  at  every  section  when  the  moving  load  covers  the  whole 
bridge. 

Funicular  polygon.  —  In  order  to  draw  the  funicular  polygon, 
it  is  practically  necessary  to  substitute  concentrated  loads  for 
the  uniformly  distributed  load.  In  determining  shear,  we  ought 
strictly  to  have  the  true  funicular  polygon  for  the  distributed 
load ;  since  without  it  we  cannot  graphically  determine  the 
amount  of  load  actually  carried  at  each  joint.  It  will,  however, 
be  sufficiently  correct  to  subdivide  the  load  into  small  portions, 
each  being  assumed  concentrated  at  its  center  of  gravity. 

The  student  will  readily  carry  out  the  construction  just  out- 
lined, every  step  being  similar  to  the  corresponding  part  of  the 
process  shown  in  Fig.  50  (PI.  IV). 

155.  Assumption  of  Equal  Panel  Loads.  — The  determination 
of  stresses,  whether  graphically  or  algebraically,  is  simplified  by 
an  approximate  assumption  which  will  now  be  explained.  This 
assumption  is  that  the  moving  load  on  the  truss  at  any  instant 
consists  of  equal  loads  concentrated  at  the  panel  joints ;  each 
load  being  equal  to  the  total  load  on  a  length  equal  to  half  the 
sum  of  the  two  adjacent  panels.  Thus,  in  Fig.  51,  in  computing 
the  shear  in  the  panel  AB,  we  assume  full  panel  loads  at  B,  C, 
D,  E,  and  F\  and  similarly  for  any  other  panel.  The  shear 
will  then  be  equal  to  the  reaction  at  A' due  to  the  series  of  loads 
to  the  right  of  the  panel  considered,  and  the  moment  curve  will 
be  simply  a  funicular  polygon  drawn  for  a  series  of  full  panel 
loads  at  all  the  joints. 

It  will  be  seen  that  the  error  involved  in  the  above  assump- 
tion is  on  the  side  of  safety,  so  far  as  the  web  members  are 


I32  GRAPHIC    STATICS. 

concerned,  since  it  gives  greater  values  of  the  shear  than 
the  more  exact  method.  In  case  of  the  bending  moments,  the 
results  are  correct  for  all  joints  of  the  loaded  chord.  For  the 
other  joints  there  will  be  a  slight  error.  This  error  will  be 
avoided,  if,  in  drawing  the  funicular  polygon,  the  load  be  con- 
sidered as  partly  supported  at  points  in  the  same  vertical  lines 
with  the  joints  of  the  unloaded  chord;  thus,  in  Fig.  51,  equal 
loads  should  be  assumed  to  act  at  all  the  points  A\  A,  B\  B,  etc. 
The  construction  just  indicated  will  not  be  here  shown,  being 
very  similar  to  that  explained  in  the  following  articles  for  the 
case  of  a  truss  with  non-parallel  chords. 

§   5.    Truss  with  Curved  Chords —  Uniform  Panel  Loads. 

156.  General  Statement.  —  If  the  upper  and  lower  chords  of 
the  truss  are  not  parallel,  the  determination  of  the  stresses  is 
somewhat  less  simple  than  in  the  case  of  parallel  chords.     But 
when  the  live  load  is  taken  as  uniformly  distributed  along  the 
bridge,  and  applied  to  the  truss  in  the  way  explained  in   Art. 
155,  the  graphic  construction  for  finding  the  maximum  stresses 
is  not  difficult ;  and  is  much  less  laborious  than  the  algebraic 
computation. 

In  Fig.  52  (PI.  V)  is  shown  a  truss  with  curved  chords, 
divided  into  equal  panels.  It  will  be  seen  that  the  method 
given  in  the  following  articles  applies  equally  to  the  case  of 
unequal  panel  lengths. 

The  principle  of  counterbracing  will  be  employed  here,  the 
diagonals  being  constructed  to  sustain  tension  only. 

157.  Dead  Load  Stresses.  —  The  dead  load  stresses  maybe 
determined  by  means  of  a  stress  diagram,  as  in  the  roof-truss 
problems  already  treated.     Two  points   should  be  observed  in 
drawing  this  diagram,     (i)   If  dead  loads  are  taken  to  act  at 
upper  as  well  as  at  lower  joints,  the  force  polygon  for  the  loads 
and  reactions  must  show  these  forces  in  the  same  order  as  that 
in  which  their  points  of  application  occur  in  the  perimeter  of 


CURVED    CHORDS  — UNIFORM   LOADS.  l^ 

the  truss.  (2)  The  diagonal  members  assumed  to  act  are 
taken  as  all  sloping  in  the  same  direction. 

The  reason  for  the  first  point  is  the  same  as  already  explained 
in  Art.  90,  —  viz.,  that  unless  the  forces  be  taken  in  the  order 
mentioned,  the  stress  diagram  cannot  be  the  true  reciprocal  of 
the  truss  diagram  and  certain  lines  will  have  to  be  duplicated. 
The  reason  for  assuming  the  diagonals  as  all  sloping  in  the 
same  way  is  the  same  as  in  the  case  of  the  roof  truss  with 
counterbracing  (Arts,  in  and  112). 

The  dead  load  stress  diagram  is  not  shown,  since  its  con- 
struction involves  no  principle  not  already  fully  explained  and 
illustrated. 

158.  Chord  Stresses  Due  to  Live  Load.  —  It  can  be  easily 
shown  that  a  load  at  any  point  of  the  truss  produces  tension  in 
every  lower  chord  member  and  compression  in  every  upper 
chord  member.  Thus,  referring  to  Fig.  52  (PI.  V),  let  a  load 
act  zkfg,  and  let  us  determine  the  kind  of  stress  caused  in  the 
member  qq\  Take  a  section  cutting  qq\  g'a',  d'd.  Consider- 
ing the  portion  of  the  truss  to  the  right  of  the  section,  the 
only  forces  acting  on  it  are  the  reaction  at  the  support  and  the 
forces  in  the  three  members  cut.  Apply  the  principle  of 
moments  to  this  system  of.  forces,  taking  the  origin  at  the 
point  of  intersection  of  g'd'  and  d'd.  It  is  evident  that  qq' 
must  be  in  compression  to  resist  the  tendency  of  the  reaction 
to  produce  left-handed  rotation  about  the  origin.  By  similar 
reasoning,  assuming  a  load  at  any  other  point,  the  student  will 
be  able  without  difficulty  to  verify  the  general  statement  above 
made. 

It  follows  that  in  order  to  get  the  greatest  possible  stresses 
in  the  chord  members,  the  truss  should  be  fully  loaded. 
Hence,  to  find  the  live-load  chord  stresses,  a  convenient 
method  is  to  draw  a  stress-diagram  for  the  truss  under  all  live 
loads.  This  diagram  is  not  shown. 


j34  GRAPHIC   STATICS. 

1 59.  Live  Load  Stresses  in  Web  Members.  —  When  the  diag- 
onals and  verticals  are  considered,  it  will  be  found  that  loads 
in  different  positions  may  cause  opposite  kinds  of  stress  in  any 
member.  Thus,  considering  the  member  f'n\  it  is  easy  to 
show  that  a  tension  is  caused  in  it  by  a  load  at  either  of  the 
points  ab,  be,  cd,  de,  or  ef,  while  a  compression  is  caused  by  a 
load  at  fg,  gh,  or  hi.  (This  may  be  shown  by  taking  a  section 
through//"',/';/,  and  n'u,  and  taking  moments  about  the  point  of 
intersection  of  the  two  chord  members  cut.)  Similarly,  loads 
at  ab,  be,  cd,  dc,  and  ef  all  tend  to  throw  compression  on  the 
vertical  member  /';;/,  while  loads  at  fg,  gh,  and  ///  have  the 
opposite  tendency. 

Therefore,  to  produce  the  greatest  tension  upon  f'n'  (and 
compression  on/';;/'),  the  live  load  must  act  only  at  ef  and  all 
joints  to  the  right ;  while  to  cause  the  greatest  compression  on 
f'n'  (and  tension  upon  /';;/),  the  live  load  must  act  only  at  fg, 
gh,  and  ///.  A  similar  statement  will  hold  regarding  any  other 
web  member.  Since  counterbraces  are  to  be  used  in  all  panels 
in  which  diagonal  members  would  otherwise  be  thrown  into 
compression,  we  shall  need  only  to  consider  the  greatest  tension 
in  each  diagonal  and  the  greatest  compression  in  each  vertical. 
We  shall  first  outline  the  method  to  be  employed,  and  then 
explain  the  construction. 

To  determine  the  greatest  tension  in  a  diagonal  member,  as 
g'm' :  Assume  the  live  load  to  come  upon  the  bridge  from  the 
right  until  there  are  full  loads  at  the  joints  ab,  be,  cd,  de,  ef,  and 
fg.  Take  a  section  cutting  g'm'  and  the  two  chord  members 
gg'  and  m'm,  and  consider  the  forces  acting  upon  the  portion  of 
the  truss  to  the  left  of  the  section.  These  forces  are  four  in 
number  :  the  reaction  at  the  support  and  the  forces  acting  in  the 
three  members  cut.  Hence  we  first  determine  the  reaction, 
and  then  determine  the  three  other  forces  for  equilibrium  by 
the  method  of  Art.  42. 

The  construction  is  shown  in  Fig.  52  (PI.  V).  ABCDEFGHI 
is  the  force  polygon  for  the  eight  live  loads  that  may  come 


CURVED   CHORDS  — UNIFORM    LOADS.  135 

upon  the  truss.  Choosing  a  pole  O,  the  funicular  polygon 
for  the  eight  loads  is  next  drawn.  Now,  turning  the  attention 
.to  the  member  g'm',  the  loads  gh  and  Id  are  assumed  not  to  act. 
The  reactions  at  the  supports  for  this  case  of  loading  are  found 
in  the  usual  way.  Prolong  oa  and  og  to  intersect  the  lines 
of  action  of  the  two  reactions,  and  join  the  two  points  thus 
determined.  This  gives  the  closing  line  of  the  funicular  poly- 
gon (or  om).  The  ray  OM  is  now  drawn  parallel  to  the  string 
om,  and  the  two  reactions  are  GMand  MA. 

Now  take  a  section  through  mm ',  m'g',  and  g'g,  and  apply 
the  construction  of  Art.  42  to  the  determination  of  the  forces 
acting  in  the  three  members  cut.  The  resultant  of  GM  and 
the  force  in  gg'  must  act  through  the  point  X  (the  intersection 
of  gg'  produced  and  ij).  The  resultant  of  the  forces  in  mm' 
and  m'g'  must  act  through  their  intersection  Y.  Hence  these 
two  resultants  (being  in  equilibrium  with  each  other)  must  both 
act  in  the  line  XY.  From  G  draw  a  line  parallel  to  gg' ,  and 
from  Ma.  line  parallel  to  X Y\  mark  their  point  of  intersection 
G'.  Then  MG'  is  the  resultant  of  the  forces  in  mm'  and  m'g'. 
From  M  draw  a  line  parallel  to  mm\  and  from  G'  a  line  parallel 
to  the  member  g'm',  and  mark  their  point  of  intersection  M'. 
Then  M'G'  represents  the  force  in  the  member  m'g'.  This  is 
also  the  value  of  the  greatest  stress  in  m'g'. 

To  determine  the  stress  in  the  vertical  member  I'g',  the 
same  loading  must  be  assumed,  and  a  similar  construction  is 
employed.  Take  a  section  cutting  II',  I'g',  and  g'g;  and  deter- 
mine forces  acting  in  these  three  lines  which  shall  be  in  equi- 
librium with  the  reaction  GM.  This  reaction  is  in  equilibrium 
with  MG'  and  G'G,  the  former  having  the  line  of  action  XY. 
Then  MG'  is  resolved  into  two  forces  having  the  directions  of 
the  members  g' I'  and  I' I.  The  stress  in  g' I'  is  found  to  be  a 
compression,  represented  in  the  stress  diagram  by  the  line  G' L' . 

The  maximum  live  load  stress  in  every  web  member  may  be 
found  in  the  same  way.  If  the  above  reasoning  is  understood, 
there  will  be  no  difficulty  in  applying  the  same  method  to  the 
remaining  members. 


136  GRAPHIC    STATICS. 

1 60.  Maximum  Stresses.  — By  combining  the  stresses  due  to 
live  and  dead  loads,  the  maximum  stresses  are  easily  deter- 
mined. 

Web  members.  —  When  the  web  members  are  considered,  the 
effect  of  counterbracing  needs  careful  attention. 

The  construction  above  explained  gives  the  greatest  live  load 
tension  in  each  diagonal.  It  may  be  that  for  certain  members, 
this  tension  is  wholly  counterbalanced  by  the  dead  loads.  In 
any  panel  in  which  this  is  the  case,  the  member  shown  will 
never  act  and  may  be  omitted.  The  counterbrace  must  then 
be  considered. 

Evidently,  the  algebraic  sum  of  the  stresses  due  to  live  and 
dead  loads  will  be  the  true  maximum  tension  in  each  of  the 
diagonals  shown.  For  the  greatest  tension  in  the  other  system 
of  diagonals,  the  load  must  be  brought  on  the  bridge  from  the 
left ;  or,  what  amounts  to  the  same  thing,  the  tension  already 
found  for  any  one  of  the  diagonals  shown  is  also  the  greatest 
tension  in  the  diagonal  sloping  the  opposite  way  in  the  panel 
equally  distant  from  the  middle  of  the  truss.  (In  fact,  the 
stresses  in  all  members  due  to  a  movement  of  the  loads  from 
left  to  right  may  be  obtained  from  the  results  already  reached, 
by  consideration  of  symmetry.) 

As  to  the  vertical  members,  two  values  of  the  stress  must  be 
compared  in  every  case  —  namely,  the  greatest  compressions 
corresponding  to  the  two  directions  of  the  moving  load.  But 
both  can  be  obtained  from  the  above  results  by  considering  the 
symmetry  of  the  truss.  For  example,  the  stress  found  for  I'g1 
is  a  possible  value  for  the  stress  in  r'b',  and  must  be  compared 
with  the  value  obtained  for  the  latter  member  when  the  load 
moves  from  right  to  left.  It  is  possible,  also,  that  certain  of 
the  verticals  may  be  in  tension  when  the  dead  loads  act  alone. 

Maximum  chord  stresses.  —  These  are  found  by  combining 
the  stresses  due  to  fixed  and  moving  loads,  determined  as 
already  described. 


CURVED   CHORDS  — CONCENTRATED    LOADS. 


137 


161.  Example. — The  following  example  should  be  carefully 
solved,  following  the  method  outlined  in  the  preceding  articles. 

Given  the  span,  120  ft.  ;  lower  chord  straight  ;  upper  chord 
circular  with  rise  at  middle  point  of  20  ft.  ;  panel  length,  15 
ft.  ;  diagonals  counterbraced  to  sustain  tension  only ;  width  of 
bridge  between  trusses,  18  ft.  Assume  dead  load  from  the 
formula  of  Art.  135,  one-third  applied  at  upper  chord  and  two- 
thirds  at  lower.  Live  load  85  Ibs.  per  square  foot.  Determine 
maximum  stresses  due  to  live  and  dead  loads. 

162.  Parallel  Chords.  —  It  will  be  noticed  that  all  the  methods 
which  have  been  described  for  the  discussion  of  trusses  with 
curved  chords  are  applicable  to  the  case  of  parallel  chords.     In 
fact,  the  determination  of  stresses  in  web  members  is  simplified 
when  both  chords  are  horizontal ;   for,   after   the    reaction    is 
found  as  in  Art.  1 59,  it  is  only  necessary  to  so  determine  the 
stress  in  the  web  member  that  its  vertical  resolved  part  shall 
equal  the  reaction. 

§  6.    Truss  with  Curved  Chords — Concentrated  Loads. 

163.  Comparison  of  Cases.  —  When  the  moving  load  consists 
of  a  series  of  unequal  weights,  the  non-parallelism  of  the  chords 
somewhat  complicates  the  determination  of  stresses.     This  case 
can,  however,  be  treated  without  great  difficulty,  as  will  now  be 
explained. 

For  the  determination  of  chord  stresses  the  method  will  be 
almost  exactly  the  same  as  in  the  case  of  parallel  chords  already 
discussed.  It  is  only  necessary  to  determine  the  greatest  bend- 
ing moment  at  each  joint,  just  as  was  done  in  Art.  150,  and 
then  find  the  chord  stresses  by  the  principle  of  moments. 

But  for  the  determination  of  web  stresses  the  shear  curve  is 
not  sufficient,  since  the  shear  in  any  section  is  not  borne  wholly 
by  the  web  member,  but  partly  by  the  inclined  chord  member. 
Moreover,  the  position  of  the  moving  load  which  will  produce 


138 


GRAPHIC   STATICS. 


the  greatest  shear  in  a  panel  is  not  generally  the  position  which 
causes  the  greatest  stress  in  the  web  member  in  that  panel. 

164.    Position  of  Loads  for  Greatest  Stress  in  a  Web  Member. 

—  We  shall  now  deduce  a  rule  for  determining  what  position  of 
the  loads  will  produce  the  greatest  stress  in  any  web  member. 
Let  XY  (Fig.  53)  represent  the  truss  and  the  member  con- 


sidered. We.  know  in  a  general  way  (Art.  1 59)  that  for  the 
greatest  tension  in  B' C  the  truss  should  be  completely  loaded 
on  the  right  of  the  panel  BC*  Let  W  =  total  load  on  truss  ; 
W  =  load  on  panel  BC\  I  =  total  length  of  truss  ;  /'  =  length 
of  panel  BC\  x  =  distance  from  Y  to  center  of  gravity  of  W\ 
x1  =  distance  from  C  to  center  of  gravity  of  W.  Prolong  the 
two  chord  members  of  the  panel  to  intersect  at  Z,  and  let 
XZ  —  a,  ZB  =  b.  Let  R  =  reaction  at  X,  and  P  =  portion  of 
W  carried  at  B.  Then 

»-?^3l§!  • 

Now  if  the  truss  be  separated  by  a  section  cutting  B1  C  and  the 
two  chord  members,  it  is  seen  that  the  moment  of  the  stress  in 
B' C  about  Z  must  equal  the  sum  of  the  moments  of  R  and  P 
about  Z.  Hence  the  stress  in  B' C  is  greatest  when  the  sum 
of  the  moments  of  R  and  P  about  Z  is  greatest.  Let  M  =  that 
sum,  then 


*  This  statement  is  true  for  all  forms  of  truss  considered  in  this  work.  If  the  two  chord 
members  in  any  panel  intersect  between  the  vertical  lines  through  the  ends  of  the  truss, 
the  statement  no  longer  holds.  The  effect  on  the  web  member  of  a  load  in  any  position 
can  be  determined  very  easily,  whatever  the  form  of  the  truss,  by  reasoning  similar  to  that 
employed  in  Art.  159. 


CURVED  CHORDS  — CONCENTRATED  LOADS. 

If  J/is  a  maximum,  we  must  have =  o.     But  since  dx'  = 

dx 

we  have, 


139 


. 
dx        II'  a    I' 

The  ratio  -  will  be  known  for  each  panel,  and  also  the  ratio      ; 
a  I 

hence  the  condition   expressed  by  the  equation   can  be  easily 
applied. 

It  may  happen  that  the  condition  is  satisfied  for  more  than 
one  position  of  the  loads.  If  this  is  so,  such  positions  must 
all  be  tried  and  the  results  compared.  This  is  illustrated  in 
the  following  article. 

Special  case.  —  It  will  be  noticed  that  if  the  chords  approach 
parallelism,  the  point  Z  moves  farther  away  and  the  limit 

approached  by   —    is  unity.     Hence,   for  the   case   of  parallel 

a  I 

chords,  the  equation  becomes    W  =  ~  W.      This    is    identical 

with  the  result  already  given  in  Art.  147. 

[NOTE.  —  The  remarks  made  in  the  note  in  Art.  147  apply  also  to  the  reasoning 
just  given.  It  is  also  to  be  noticed  that  we  have  assumed  that  no  load  is  between 
B  and  X.  If  the  condition  above  deduced  cannot  be  satisfied  without  carrying  the 
foremost  load  or  loads  to  the  left  of  B,  we  may  reason  thus :  Let  W"  =  load  to  left 
of  B;  x"  =  distance  of  center  of  gravity  of  W"  from  B,  Then 

M-  Ra-  Pb-  W"  (b  -  *")  =  — 

Differentiating,  remembering  that  Jx"  =  dx'  —  dx,  we  have 

w,, 
dx  I  V 

Placing  this  equal  to  zero,  we  have 

w—'L*L  iv1  -  -  iv". 

a      V  a 

Usually  W'1  will  be  zero  and  the  equation  first  deduced  will  apply,  but  if  necessary 

the  last  equation  may  be   used.     For  the  case  of  parallel  chords,  the  term  -  W 
I  & 

disappears,  since  -  approaches  zero.] 
a 


140 


GRAPHIC   STATICS. 


165.  Determination  of  Web  Stresses. — The  method  of  deter- 
mining web  stresses  for  the  case  now  under  consideration  is 
shown  in  Fig.  54  (PI.  V).  The  force  polygon  and  funicular 
polygon  for  the  series  of  wheel  loads  is  drawn  in  the  usual 
manner.  The  span  of  the  truss  here  taken  is  120  ft.,  divided 
into  eight  equal  panels,  the  upper  chord  being  curved,  while 
the  lower  (supporting  the  floor  system  and  moving  loads)  is 
straight.  Half  of  the  truss  is  shown  at  XY,  drawn  to  the 
same  scale  used  in  the  space  diagram  for  the  moving  loads. 
The  construction  is  shown  for  finding  stresses  in  the  two  mem- 
bers tu  and  uv.  First  consider  the  latter. 

To  apply  the  rule  of  the  preceding  article  for  finding  the 

required  position  of  loads,  determine  the  ratio  -.      Prolonging 

a 

the  two  chord  members  ku  and  vm  till  they  intersect  at  Zz,  we 

find  —££.=— =2.      Also  we  have  -  =  8.      Hence,  for  the  great- 
Z%X    a  I 

est  stress  in  uv,  the  load  on  the  panel  QR  must  equal  one- 
sixteenth  the  load  on  the  whole  truss.  Now  it  is  easily  seen 
that  this  condition  is  satisfied  when  the  second  load  is  at  the 
point  R.  For,  in  this  position,  the  whole  load  on  the  truss  is 
123750  Ibs.  If  the  second  load  is  just  at  the  left  of  R,  the 
load  on  the  panel  is  20000  Ibs.,  which  is  greater  than  one- 
sixteenth  of  1 23750  Ibs.  ;  but  if  the  second  load  is  just  at  the 
right  of  R,  the  load  on  the  panel  is  7500  Ibs.,  which  is  less  than 
one-sixteenth  of  123750  Ibs. 

We  are  now  ready  to  determine  the  stress  in  the  member  uv. 
The  general  method  is  to-  cut  the  truss  by  a  section  through 
ku,  uv,  and  vm ;  determine  the  resultant  of  all  forces  on  the 
truss  to  the  left  of  the  section ;  and  then  determine  three 
forces  in  the  members  cut  which  shall  be  in  equilibrium  with 
that  resultant. 

The  forces  acting  on  the  truss  to  the  left  of  the  section  are 
the  reaction  at  the  support  and  the  portion  of  the  foremost 
load  AB  carried  at  the  point  Q.  Their  resultant  is  found  as 
follows :  Draw  XUX«  to  represent  the  span  when  the  load  BC  is 


CURVED    CHORDS  — CONCENTRATED    LOADS.  i4I 

at  R  ;  then  draw  the  closing  line  of  the  funicular  polygon  for 
the  forces  now  on  the  truss,  and  from  O  draw  the  ray  parallel 
to  this  closing  line.  The  end  of  this  ray  is  marked  MI,  and  the 
left  reaction  is  represented  by  the  line  M\A.  Next  replace  AB 
by  its  two  components  borne  at  Q  and  R.  These  are  deter- 
mined by  drawing  verticals  from  Q  and  R  intersecting  the 
strings  oa  and  ob  in  Q'  and  R! ;  then  Q'R'  is  the  string  which 
must  replace  oa  and  ob.  (The  portions  of  the  funicular  poly- 
gon to  the  right  of  R'  and  to  the  left  of  Q'  are  not  changed  by 
thus  replacing  AB  by  its  two  components  at  Q  and  R.)  Now 
by  drawing  from  O  a  ray  parallel  to  Q'R'  we  find  the  point 
(marked  Kj)  which  divides  AB  into  the  two  components  at  Q 
and  R. 

The  resultant  of  the  reaction  at  X  and  the  load  at  Q  is  given 
in  magnitude  by  M\K±.  We  need  also  its  line  of  action,  which 
is  found  as  follows  :  Prolong  Q'R'  and  om\  (the  closing  string  of 
the  funicular  polygon  for  all  the  forces  on  the  truss)  until  they 
intersect  in  z' ;  the  required  line  of  action  is  a  vertical  line 
through  s'.  This  vertical  line  intersects  XUXV  produced  in  z. 
We  now  locate  the  point  z  in  the  truss  diagram  by  making 
Xz  =  Xuz  in  the  diagram  above. 

We  have  now  to  solve  the  following  problem  :  Determine 
three  forces  acting  in  lines  ku,  uv,  and  vm,  which  shall  be  in 
equilibrium  with  a  force  equal  to  the  resultant  just  found  and 
acting  upward  in  a  line  through  z.  This  is  solved  by  the 
method  of  Art.  42.  Draw  KM  equal  to  the  given  force  acting 
at  s.  Draw  zQ\  and  determine  two  forces,  one  parallel  to 
zQ  and  the  other  to  kit,  which  shall  be  in  equilibrium  with 
KM.  This  gives  UK  as  the  force  in  the  line  uk  and  MU  as 
the  resultant  of  the  forces  in  the  lines  mv  and  vu.  From  M 
and  U  draw  lines  parallel  to  mv  and  vu  respectively,  and  mark 
their  intersection  V;  then  J/Fand  VU  represent  the  stresses 
in  the  corresponding  truss  members.  The  latter  is  evidently 
the  required  stress  in  the  member  uv. 

Turning  next  to  the  member  ///,  a  section  through  it  will  cut 


1 42 


GRAPHIC    STATICS. 


the  two  chord  members  kit  and  mt ;  these  intersect  at  Z±     By 

measurement  we  find  -  =  ^-^=6,  and =  8x6  =  48.     There- 

a     Z^X  I    a 

fore,  for  the  greatest  stress  in  tu  we  must  have  the  load  on  the 
panel  QR  equal  to  the  whole  load  on  the  truss  divided  by  48. 
This  condition  is  seen  to  be  satisfied  when  the  foremost  load  is 
at  R.  Since  in  this  case  there  is  no  load  borne  to  the  left  of 
R,  the  only  force  acting  on  the  truss  to  the  left  of  a  section 
through  tu  is  the  reaction  at  the  support.  To  find  this,  draw 
XtXu  to  represent  the  length  of  the  truss  for  the  required  posi- 
tion of  the  loads,  and  find  om\>  the  closing  string  of  the  funic- 
ular polygon  for  all  loads  and  reactions  on  the  truss.  The 
corresponding  ray  is  OMJ,  and  M^A  represents  the  reaction 
at  the  left  support. 

Referring  now  to  the  truss  diagram,  we  have  to  determine 
forces  acting  in  the  three  lines  mt,  tu,  uk,  which  shall  be  in 
equilibrium  with  the  reaction  just  determined.  Lay  off  JCMto 
represent  this  reaction  ;  draw  a  line  from  K'  parallel  to  ku  and 
one  from  M  parallel  to  Q'X,  and  mark  their  point  of  intersec- 
tion U' ;  then  [7'K1  represents  the  force  in  uk,  and  MU'  the 
resultant  of  the  forces  in  mt  and  tu.  From  M  draw  a  line  par- 
allel to  mt  and  from  [/'  a  line  parallel  to  ut,  intersecting  at  T; 
then  TU1  represents  the  required  stress  in  tu. 

The  value  just  determined  is  the  true  maximum  stress  for 
the  member  /;/.  In  case  of  the  member  uv,  however,  further 
investigation  is  needed.  It  will  be  found  by  trial  that  the 

condition  W  =  -  •  —  •   W  is  satisfied  for  the  member  uv,  not 
a      I' 

only  when  the  loads  have  the  position  already  treated,  but  also 
when  the  foremost  load  is  at  R.  Hence  the  stress  in  uv  for 
this  position  of  the  loads  must  be  determined  and  compared 
with  the  one  already  found,  namely  UV.  For  this  case  of 
loading  the  only  force  acting  on  the  truss  to  the  left  of  a  section 
through  uv  is  the  reaction  (already  found  when  considering  the 
member  tu)  represented  by  K'M.  Making  this  in  equilibrium 
with  three  forces  whose  lines  of  action  are  kn,  uv,  vm,  we  find 


CURVED  CHORDS  — CONCENTRATED  LOADS.     143 

U'  I7'  as  the  stress  in  uv.  Comparison  shows  that  this  is 
slightly  greater  than  UV\  hence  the  value  last  found  is  the 
true  maximum  for  the  member  nv. 

The  construction  for  each  of  the  other  web  members  is 
exactly  similar.  If  both  chord  members  in  any  panel  are 
inclined,  the  construction  requires  only  slight  modification, 
which  the  student  will  readily  supply. 

In  case  the  intersection  of  two  chord  members  cannot  con- 
veniently be  found  by  prolonging  them,  the  distances  a  and  b 
can  be  easily  computed  from  the  dimensions  of  the  truss. 

Remark.  —  It  will  often  be  found  that  two  web  members 
related  as  are  tu  and  nv,  will  not  sustain  their  maximum  stresses 
at  the  same  time.  In  the  case  of  parallel  chords  this  could  not 
occur. 

1 66.  Minimum  Stresses.  —  In  what  precedes,  little  has  been 
said  of  minimum  stresses  in  the  truss  members.  If,  however, 
the  design  is  to  be  made  in  accordance  with  the  theory  of  the 
strength  of  materials  under  repeated  alternations  of  stress,  the 
minimum  stress  sustained  by  each  member  becomes  important. 
In  all  the  cases  treated  in  the  present  chapter,  the  minimum 
stresses  can  be  determined  without  difficulty,  without  the  use 
of  additional  principles. 


PART   III. 


CENTROIDS  AND  MOMENTS  OF  INERTIA. 


CHAPTER   VIII.      CENTROIDS. 
§  I.    Centroid  of  Parallel  Forces. 

167.  Composition  of  Parallel  Forces. — The   composition   of 
complanar  parallel  forces    can    always    be   effected   by   means 
of  the  funicular  polygon,  by  the  method  of  Art.  27.     It  is  now 
necessary  to  consider  parallel  systems  more  at  length,  as  a  pre- 
liminary to  the  discussion  of  graphic  methods  for  determining 
centers  of  gravity  and  moments  of  inertia. 

168.  Resultant  of  Two  Parallel  Forces.  —Let  ab  and  be  (Fig. 
55)  be  the  lines  of  action  of  two  parallel  forces,  their  magni- 
tudes being  AB  and  BC  (not  shown).     Let 
ac  be  the  line  of  action  of  their  resultant, 
and  AC  (not  shown)   its    magnitude.       By 
the  principle  of  moments  (Art.  50)  the  sum 
of  the  moments  of  AB  and  BC  about  any 

Fig.  ss  point  in  their  plane  is  equal  to  the  moment 

of  A C  about  the  same  point.  If  the  origin  of  moments  is  on 
ac,  the  moment  of  AC  is  zero;  and  therefore  the  moments  of 
AB  and  BC  are  numerically  equal  (but  of  opposite  signs). 

Let  any  line  be  drawn  perpendicular  to  the  given  forces, 
intersecting  their  lines  of  action  in  P',  Q',  and  R'  respectively. 


ab 
P 


144 


CENTROID  OF  PARALLEL  FORCES. 


Let  any  other  line  be  drawn  intersecting  the  three  lines  of 
action  in  P,  Q,  and  R  respectively.  Then 

PR        QR 
P'R'     Q'R'* 

and  therefore  ABxPR=BCx  QR. 

That  is,  PQ  is  divided  by  the  line  ac  into  segments  inversely 
proportional  to  AB  and  BC. 

If  AB  and  BC  act  in  opposite  directions,  the  line  ac  will  be 
outside  the  space  included  between  ab  and  be ;  but  the  above 
result  is  true  for  either  case. 

169.  Centroid  of  Two  Parallel  Forces.  — If  the  lines  of  action 
ab  and  be  (Fig.  55)  be  turned  through  any  angle  about  the  points 
P  and   Q  respectively,    the  forces   remaining  parallel   and    of 
unchanged  magnitudes,  the  line  of  action  of  their  resultant  will 
always  pass  through  the  point  R.     For,  by  the  preceding  article, 
the  line  of  action  of  the  resultant  will  always  intersect  PQ  in  a 
point  which  divides  PQ  into  segments  inversely  proportional  to 
AB  and  BC.     Hence,  if  AB  and  BC  remain  unchanged,  and 
also  the  points  P  and  Q,  the  point  R  must  also  remain  fixed. 

If  P  and  Q  are  taken  as  the  points  of  application  of  AB  and 
BC,  R  may  be  taken  as  the  point  of  application  of  AC,  in 
whatever  direction  the  parallel  forces  are  supposed  to  act.  The 
point  R  is  called  the  centroid  *  of  the  parallel  forces  AB  and 
f>C  for  the  fixed  points  of  application  P  and  Q. 

170.  Centroid  of  Any  Number  of  Parallel  Forces.  —  Let  AB, 

BC,   and    CD  be  three  parallel  forces,  and  let   P,  Q,  and   S 

(Fig.   56)  be  their  fixed  points  of 

application.      Let    R  be  the  cen- 

troid  of  AB  and  BC,  and  A  C  their 

resultant.     Take   R   as    the    fixed 

point    of   application    of   AC,   and 

determine   Tt  the  centroid  of  AC  Fig.  se 

*  The  name  center  of  parallel  forces  has  been  quite  commonly  used  instead  of  centroid 
as  above  defined.  The  latter  term  has,  however,  been  adopted  by  some  of  the  later  writers, 
and  seems  on  the  whole  a  better  designation. 


a  &      a 


I46  GRAPHIC   STATICS. 

and  CD.  Let  AD  be  the  resultant  of  AC  and  CD ;  then  AD  is 
also  the  resultant  of  ABt  BCt  and  CD. 

Now  if  AB,  BC,  and  CD  have  their  direction  changed,  but 
still  remain  parallel  and  unchanged  in  magnitude,  it  is  evident 
that  the  point  T,  determined  as  above,  will  remain  fixed  and  will 
always  be  on  ad,  the  line  of  action  of  AD.  The  point  T is  called 
the  centroid  of  the  three  forces  AB,  BC,  and  CD. 

By  an  extension  of  the  same  method,  a  centroid  may  be 
determined  for  any  system  of  parallel  forces  having  fixed  points 
of  application.  Hence  the  following  definition  may  be  stated  : 

The  centroid  of  a  system  of  parallel  forces  having  fixed  points 
of  application  is  a  point  through  which  the  line  of  action  of 
their  resultant  passes,  in  whatever  direction  the  forces  be  taken 
to  act. 

In  determining  the  centroid  by  the  method  just  described, 
the  forces  may  be  taken  in  any  order  without  changing  the 
results.  For  the  centroid  must  lie  on  the  line  of  action  of  the 
resultant ;  and  since  this  is  a  determinate  line  for  each  direction 
in  which  the  forces  may  be  taken  to  act,  there  can  be  but  one 
centroid. 

171.  Non-complanar  Parallel  Forces.  —  The  reasoning  of  the 
preceding  articles  is  equally  true,  whether  the  forces  are  corn- 
planar   or   not.      In   what   follows   we    shall   deal   either    with 
complanar  forces,  or  with  forces  whose  points  of  application  are 
complanar.     No  more  general  case  will  be  discussed. 

172.  Graphic  Determination  of  Centroid  of  Parallel  Forces.— 

If  the  line  of  action  of  the  resultant  of  any  system  of  parallel 
forces  be  found  for  each  of  two  assumed  directions  of  the  forces, 
the  point  of  intersection  of  these  two  lines  is  the  centroid  of 
the  system.  Moreover,  if  the  points  of  application  are  com- 
planar, the  two  assumed  directions  may  both  be  such  that  the 
forces  will  be  complanar. 

Thus,  let  the  plane  of  the  paper  be  the  plane  containing  the 
given  points  of  application,  and  let  ab,  be,  cd,  de,  ef  (Fig.  57)  be 


CENTROID  OF  PARALLEL  FORCES.         l^ 

the  points  of  application  of  five  parallel  forces,  AB,  BC,  CD, 
DE,  EF.  Draw  through  these  points  parallel  lines  in  some 
chosen  direction,  and  taking  them  as  the  lines  of  action  of  the 
given  forces,  construct  the  force  and  funicular  polygons  corre- 


sponding.  The  line  of  action  of  the  resultant  is  drawn  through 
the  intersection  of  the  strings  oa  and  of,  and  this  line  contains 
the  centroid  of  the  given  forces.  Next  draw  through  the  given 
points  of  application  another  set  of  parallel  lines,  preferably  per- 
pendicular to  the  set  first  drawn,  and  draw  a  funicular  polygon 
for  the  given  forces  with  these  lines  of  action.  (It  is  unneces- 
sary to  draw  a  new  force  diagram,  since  the  strings  in  the 
second  funicular  polygon  may  be  drawn  respectively  perpendic- 
ular to  those  of  the  first.)  This  construction  determines  a 
second  line  as  the  line  of  action  of  the  resultant  corresponding 
to  the  second  direction  of  the  forces.  The  required  centroid  of 
the  system  is  the  point  of  intersection  of  the  two  lines  of 
action  of  the  resultant  thus  determined,  and  is  the  point  af  in 
the  figure. 

Example.  —  Find  graphically  the  centroid  of  the  following 
system  of  parallel  forces,  and  test  the  result  by  algebraic  com- 
putation :  A  force  of  20  Ibs.  applied  at  a  point  whose  rectangu- 
lar coordinates  are  (4,  6)  ;  12  Ibs.  at  the  point  (12,  3)  ;  20  Ibs.  at 


T48  GRAPHIC    STATICS. 

the  point  (10,  10)  ;   —  10  Ibs.  at  the  point  (7,  —9);   —8  Ibs.  at 
the  point  (—5,  —  10). 

173.  Centroid  of  a  Couple.  —  If  AB  and  BC  are  the  magni- 
tudes of  any  two  parallel  forces  applied  at  points  P  and  Q, 
then  by  Art.  169  their  centroid  R  is  on  the  line  PQ  and  is 
determined  by  the  equation 


QR     AB' 

If  AB  and  BC  are  equal  and  opposite  forces,  PR  and  QR  must 
be  numerically  equal,  and  R  must  be  outside  the  space  between 
the  lines  of  action  of  AB  and  BC.  These  conditions  can  be 
satisfied  only  by  making  PR  and  QR  infinite.  Hence  we  may 
say  that  the  centroid  of  two  equal  and  opposite  forces  lies  on 
the  line  joining  their  points  of  application  'and  is  infinitely  dis- 
tant from  these  points. 

174.    Centroid  of  a  System   whose  Resultant  is  a  Couple.— 

If  a  system  with  fixed  points  of  application  is  equivalent  to  a 
couple,  its  centroid  will  be  infinitely  distant  from  the  given 
points  of  application.  A  line  containing  this  centroid  can  be 
determined  as  follows  : 

Take  the  given  forces  in  two  groups  ;  the  resultants  of  the 
two  groups  will  be  equal  and  opposite.  Find  the  centroid  of 
each  group,  and  suppose  each  partial  resultant  applied  at  the 
corresponding  centroid.  Then  the  centroid  of  the  whole  sys- 
tem is  the  same  as  that  of  the  couple  thus  formed,  and  will  lie 
on  the  line  joining  the  two'  partial  centroids. 

If  the  separation  into  groups  be  made  in  different  ways,  dif- 
ferent couples  and  different  partial  centroids  will  be  found. 
The  different  couples  are,  of  course,  equivalent  ;  and  it  may  be 
proved  that  the  lines  joining  the  different  pairs  of  partial  cen- 
troids are  all  parallel,  and  intersect  in  the  (infinitely  distant) 
centroid  of  the  whole  system. 

For,  suppose  the  given   system  to  be  equivalent  to  a  couple 


CENTER   OF   GRAVITY. 


149 


Q  with  points  of  application  A  and  B,  and  also  to  a  couple  Q' 
with  points  of  application  A'  and  />'.  These  two  couples  must 
be  equivalent  to  each  other,  whatever  be  the  direction  of  the 
forces.  Let  AB  be  taken  as  this  direction.  The  two  equal 
and  opposite  forces  of  the  couple  Q  neutralize  each  other,  since 
their  lines  of  action  are  coincident ;  hence  the  two  forces  of 
the  couple  Q  must  also  neutralize  each  other.  Therefore  A'£' 
must  be  parallel  to  AB. 

175.  Moment  of  a  Force  about  an  Axis.  —  The  moment  of  a 
force  with  respect  to  a  given  axis,  as  defined  in  Art.  47, 
depends  not  only  upon  the  point  of  application  of  the  force, 
but  upon  its  direction.  In  dealing  with  systems  of  forces 
whose  direction  may  change  but  whose  points  of  application 
are  complanar,  we  shall  need  to  compute  moments  only  for 
axes  lying  in  the  plane  of  the  points  of  application  ;  and  the 
forces  may  usually  be  regarded  as  acting  in  lines  perpendicular 
to  this  plane.  Hence  we  shall  compute  moments  by  the 
following  rule  : 

The  moment  of  a  force  with  reference  to  any  axis  is  the 
product  of  the  magnitude  of  the  force  into  the  distance  of  its 
point  of  application  from  the  axis.  . 


§   2.    Center  of  Gravity  —  Definitions  and   General  Principles. 

176.  Center  of  Gravity  of  Any  Body.  —  Every  particle  of 
a  terrestrial  body  is  attracted  by  the  earth  with  a  force  propor- 
tional directly  to  the  mass  of  the  particle.  The  resultant  of 
such  forces  upon  all  the  particles  of  a  body  is  its  iveight ;  and 
the  point  of  application  of  this  resultant  is  called  the  center  of 
gravity  of  the  body.  The  lines  of  action  of  these  forces  may 
be  assumed  parallel  without  appreciable  error.  We  may  there- 
fore define  the  center  of  gravity  of  a  body  as  follows  : 

If  forces  be  supposed  to  act  in  the  same  direction  upon  all 
particles  of  a  body,  each  force  being  proportional  to  the  mass 


150  GRAPHIC    STATICS. 

of  the  particle  upon  which  it  acts,  the  centroid  of  this  system 
of  parallel  forces  is  the  center  of  gravity  of  the  body. 

This  point  is  also  called  center  of  mass,  and  center  of  inertia, 
either  of  which  is  a  better  designation  than  center  of  gravity. 
The  latter  term  is,  however,  in  more  general  use. 

177.  Centers   of   Gravity  of   Areas   and   Lines.  —  The   term 
center  of    gravity,   as   above  defined,   has    no    meaning  when 
applied   to   lines  and  areas,   since   these  magnitudes  have  no 
mass,  and  hence  are  not  acted  upon  by  the  force  of  gravity. 
It  is,  however,  common  to  use  the  name  center  of  gravity  in 
the  case  of  lines  and  areas,  with  meanings  which  may  be  stated 
as  follows  : 

The  center  of  gravity  of  an  area  is  the  center  of  gravity  of 
its  mass,  on  the  supposition  that  each  superficial  element  has  a 
mass  proportional  to  its  area.  This  point  would  be  better 
described  as  the  center  of  area. 

The  center  of  gravity  of  a  line  is  the  center  of  gravity  of  its 
mass,  on  the  assumption  that  each  linear  element  has  a  mass 
proportional  to  its  length.  The  term  center  of  length  is  pref- 
erable, and  will  often  be  used  in  what  follows. 

Similar  statements  might  be  made  regarding  geometrical 
solids,  but  we  shall  have  to  deal  chiefly  with  lines  and  areas. 

178.  Moments  of  Areas  and   Lines. — Definition. — The  mo- 
ment of  a  plane  area  with  reference  to  an  axis  lying  in  its  plane 
is  the  product  of    the  area  by  the  distance  of  its  center  of 
gravity  from  the  axis. 

Proposition.  —  The  moment  of  any  area  about  a  given  axis  is 
equal  to  the  sum  of  the  moments  of  any  set  of  partial  areas 
into  which  it  may  be  divided.  For,  by  the  definition  of  center 
of  gravity,  a  force  numerically  equal  to  the  total  area  and 
applied  at  its  center  of  gravity  is  the  resultant  of  a  system  of 
forces  numerically  equal  to  the  partial  areas  and  applied  at 
their  respective  centers  of  gravity  ;  and  the  moment  of  any 


CENTER   OF   GRAVITY.  !5! 

force  is  equal  to  the  sum  of  the  moments  of  its  components 
(Art.  50). 

A  similar  definition  and  proposition  may  be  stated  regarding 
lines. 

The  moment  of  an  area  or  line  is  zero  for  any  axis  containing 
its  center  of  gravity. 

179.  Symmetry.  —  Two   points   are    symmetrically  situated 
with  respect  to  a  third  point  if  the  line  joining  them  is  bisected 
by  that  point. 

Two  points  are  symmetrically  situated  with  respect  to  a  line 
or  plane  when  the  line  joining  them  is  perpendicular  to  the 
given  line  or  plane  and  bisected  by  it. 

A  body  is  symmetrical  with  respect  to  a  point,  a  line,  or  a 
plane,  if  for  every  point  in  the  body  there  is  another  such  that 
the  two  are  symmetrically  situated  with  respect  to  the  given 
point,  line,  or  plane.  The  point,  line,  or  plane  is  in  this  case 
called  a  point  of  symmetry,  an  axis  of  symmetry,  or  a  plane  of 
symmetry  of  the  body. 

1 80.  General  Principles.  —  (i)  The  center  of  gravity  of  two 
masses  taken  together  is  on   the  line  joining  the   centers  of 
gravity  of  the  separate  masses.     For  it  is  the  point  of  applica- 
tion of   the  resultant  of  two  parallel  forces  applied  at  those 
points. 

(2)  If  a  body  of  uniform  density  has  a  plane  of  symmetry, 
the  center  of  gravity  lies  in  this  plane.  If  there  is  an  axis  of 
symmetry,  the  center  of  gravity  lies  in  this  axis.  If  the  body 
is  symmetrical  with  respect  to  a  point,  that  point  is  the  center 
of  gravity.  For  the  elementary  portions  of  the  body  may  be 
taken  in  pairs  such  that  for  each  pair  the  center  of  gravity  is 
in  the  plane,  axis,  or  point  of  symmetry. 

181.  Centroid. — The  center  of  gravity  of  any  body  or  geo- 
metrical magnitude  is  by  definition  the  same  as  the  centroid  of 
a   certain    system    of   parallel   forces.     It    will    be    convenient, 


152 


GRAPHIC    STATICS. 


therefore,  to  use   the  word   centroid  in  most  cases  instead  of 
center  of  gravity. 

§   3.    Centroids  of  Lines  and  of  Areas. 

182.  General   Method   of   Finding   Centroid.  —  The  centroid 
of  any  area  may  be  found  by  the  following  method  :   Divide  the 
given  area  into  parts  such  that  the  area  and  centroid  of  each 
part  are  known.     Take  the  centroids  of  the  partial  areas  as  the 
points  of  application  of  forces  proportional  respectively  to  those 
areas.     The  centroid  of  this  system  of  forces  is  the  centroid  of 
the  total  area,  and  may  be  found  by.  the  method  of  Art.  172. 

The  centroid  of  a  line  may  be  found  by  a  similar  method. 

The  method  just  described  is  exact  if  the  magnitudes  of  the 
partial  areas  and  their  centroids  are  accurately  known.  If  the 
given  area  is  such  that  it  cannot  be  divided  into  known  parts,  it 
will  still  be  possible  to  get  an  approximate  result  by  this 
method. 

In  applying  this  general  method,  it  is  frequently  necessary  to 
know  the  centroids  of  certain  geometrical  lines  and  figures,  and 
also  the  relative  magnitudes  of  the  areas  of  such  figures. 
Methods  of  determining  these  will  be  given  in  the  following 
articles. 

183.  Centroids  of  Lines. — The  centroid  of  a  straight  line  is 
at  its  middle  point. 

Broken  line.  —  The  centroid  of  a  broken  line  is  the  center  of 
a  system  of  parallel  forces,  of  magnitudes  proportional  to  the 
lengths  of  the  straight  portions  of  the  line,  and  applied  respec- 
tively at  their  middle  points.  It  may  be  found  graphically  by 
the  method  of  Art.  172,  or  by  any  other  method  applicable  to 
parallel  forces. 

Part  of  regular  polygon.  —  For  the  centroid  of  a  part  of  a 
regular  polygon,  a  special  construction  is  found  useful. 


CENTROIDS    OF    LINES    AND    OF   AREAS. 


153 


Let  ABODE  (Fig.  58)  be  part  of  such  a  polygon,  and  O  the 

center  of  the  inscribed  circle.  Let 
r  —  radius  of  inscribed  circle  ;  /  = 
length  of  a  side  of  the  polygon  ;  s 
=  total  length  of  the  broken  line 

I    .         xJ  _  AE.     Through  O  draw  OO,  the  axis 

M  A'Q  E/N        of  symmetry  of  AE]  and  MN,  per- 

pendicular to  OC. 
First,  the  centroid  must  lie  on  OC. 

Second,  to  find  its  distance  from  O,  assume  a  system  of  equal 
and  parallel  forces  applied  at  the  middle  points  of  the  sides  AB, 
BO,  etc.  The  required  centroid  is  the  point  of  application  of 
the  resultant  of  these  forces.  Taking  MN  as  the  axis  of 
moments,  and  letting  x  =  required  distance  of  centroid  from 
MN,  and  x^  x2,  xz,  x^  the  distances  of  the  middle  points  of  AB, 
BC,  etc.,  from  MN,  we  have  from  the  principle  of  moments, 

/X]_  4-  fat  +  lx%  +  lx±  =  sx. 

But  if  Ab  and  Bb  be  drawn  perpendicular  respectively  to  PQ 
and  MJVwQ  have  from  the  similar  triangles  ABb,  POQ. 


Ab-    PQ      Ab 

.:  Ix^r-Ab. 
In  the  same  way, 

lxz  =  r-  be, 


Ix^r-dE. 
Hence,  s*x=r  (Ab  +  bc  +  cd+dE)  =  r  •  AE, 

where  AE  is  equal  to  the  projection  of  the  broken  line  ABODE 
on  MN. 

The  centroid  G  may  now  be  found  graphically  as  follows  : 
Make6fc'  =  r;  ON=\s\  OE'  =  ^AE;  draw  Nc'  .  Then  G  is 
determined  by  drawing  E'G  parallel  to  Nc'. 


154  GRAPHIC   STATICS. 

Circular  arc.  —  The  above  construction  holds,  whatever  the 
length  of  the  side  /.  If  this  length  be  decreased  indefinitely, 
while  the  number  of  sides  is  increased  indefinitely,  so  that  the 
length  s  remains  finite,  we  reach  as  the  limiting  case  a  circular 
arc.  The  same  construction  therefore  applies  to  the  determina- 
tion of  the  centroid  of  such  an  arc,  r  denoting  the  radius  of  the. 
circle  and  s  the  length  of  the  arc. 

184.  Centroids  of  Geometrical  Areas.  —  Parallelogram.  —  The 
centroid  of  a  parallelogram  is  on  a  line  bisecting  two  opposite 
sides. 

Let  ABCD  (Fig.  59)  be  a  parallelogram,  and  EF  a  line 
bisecting  AD  and  BC.  Divide  AB  into  any  even  number  of 

equal  parts,  and  through  the  points 
of  division  draw  lines  parallel  to  BC. 
Also  divide  BC  into  any  even  number 
of  equal  parts  and  draw  through  the 
points  of  division  lines  parallel  to 
AB.  The  given  parallelogram  is  thus 
divided  into  equal  elements.  Now 
consider  a  pair  of  these"  elements,  such  as  those  marked  P  and 
Q  in  the  figure,  equally  distant  from  AD,  and  also  equally 
distant  from  EF,  but  on  opposite  sides  of  it.  The  centroid  of 
the  two  elements  taken  together  is  at  the  middle  point  of  the 
line  joining  their  separate  centroids.  If  the  number  of  divisions 
of  AB  and  of  BC  be  increased  without  limit,  the  elements 
approach  zero  in  area,  and  the  centroids  of  P  and  Q  evidently 
approach  two  points  which  are  equally  distant  from  EF.  Hence 
in  the  limit,  the  centroid  of  such  a  pair  of  elements  lies  on  the 
line  EF.  But  the  whole  area  ABCD  is  made  up  of  such  pairs  ; 
hence  the  centroid  of  the  whole  area  is  on  the  line  EF.  For 
like  reasons  it  is  also  on  the  line  bisecting  AB  and  DC',  hence 
it  is  at  the  intersection  of  the  two  bisectors. 

The  point  thus  determined  evidently  coincides  with  the  point 
of  intersection  of  the  diagonals  AC  and  BD. 


CENTROIDS    OF    LINES   AND   OF   AREAS. 


55 


Triangle.  —  The  centroid  of  a  triangle  lies  on  a  line  drawn 
from  any  vertex  to  the  middle  of  the  opposite  side ;  and  is, 
therefore,  the  point  of  intersection  of  the  three  such  lines. 

Let  ABC  (Fig.  60)  be  any  triangle,  and  D  the  middle  point 
of  BC.  Then  the  centroid  of  ABC  must  lie  on  AD.  For  AD 
bisects  all  lines,  such  as  be,  parallel  to  BC.  Now  inscribe  in 
the  triangle  any  number  of  parallelograms  such  as  bcc*b\  with 
sides  parallel  respectively  to  BC  and 
AD.  The  centroid  of  each  parallelo- 
gram lies  on  AD,  and,  therefore,  so 
also  does  the  centroid  of  the  whole 
area  composed  of  such  parallelograms. 
If  the  number  of  such  parallelograms 
be  increased  without  limit,  the  alti- 
tude of  each  being  diminished  without  limit,  their  combined 
area  will  approach  that  of  the  triangle,  and  the  centroid  of  this 
area  will  approach  in  position  that  of  the  triangle.  But  since 
the  former  point  is  always  on  the  line  AD,  its  limiting  position 
must  be  on  that  line.  Therefore  the  line  AD  contains  the  cen- 
troid of  the  triangle. 

By  the  same  reasoning,  it  follows  that  the  centroid  of  ABC 
must  lie  on  BE,  drawn  from  B  to  the  middle  point  of  AC. 
Hence  it  must  be  the  point  of  intersection  of  AD  and  BE, 
which  point  must  also  lie  on  the  line  CF  drawn  from  C  to  the 
middle  point  of  AB. 

The  point  G  divides  each  bisector  into  segments  which  are 
to  each  other  as  i  to  2.  For,  from  the  similar  triangles  ABC, 
EDC,  since  EC  is  half  of  AC,  it  follows  that  DE  is  equal  to 
half  of  BA.  And  from  the  similar  triangles  AGB,  DGE,  since 
DE  is  half  of  AB,  it  follows  that  GE  is  half  of  GB,  and  GD 
half  of  GA. 

Quadrilateral.  —  Let  ABCD  (Fig.  61)  be  a  quadrilateral  of 
which  it  is  required  to  find  the  center  of  gravity.  Draw  BD, 
and  let  E  be  its  middle  point.  Make  EGi  =  J  EA,  and  EG2=\ 
EC.  Then  the  centroids  of  the  triangles  ABD  and  BCD  are 


1 56 


GRAPHIC   STATICS. 


Fig.  61 


GI  and  G-2  respectively.     Hence  the  centroid  of  ABCD  is  on 
the  line  GiG2  at   a  point  dividing  it    into  segments  inversely 

proportional  to  the  areas  of  ABD 
and  BCD.  Since  these  two  tri- 
angles have  a  common  base,  their 
areas  are  proportional  to  their  alti- 
tudes measured  from  this  base. 
But  these  altitudes  are  propor- 
tional to  AF  and  FC,  or  to  GJi  and  G*H ;  hence,  if  G  is  the 
required  centroid, 

G\G  '.  G->G  '. '.  G-iH '.  G\f~f. 

Therefore  G  is  found  by  making  G1G=G2H. 

Circular  sector. — To  find  the  centroid  of  a  circular  sector 
OAB  (Fig.  62)  we  may  reason  as  follows :  Draw  two  radii  OM, 
ON,  very  near  together.  Then  OMN  differs  little  from  a  tri- 
angle, and  its  centroid  will  fall  very  near  the  arc  AB\  drawn 
with  radius  equal  to  two-thirds  of  OA.  If  the  whole  sector  be 
subdivided  into  elements  such  as  OMN,  their  centers  of  gravity 
will  all  fall  very  near  to  the  arc  AB\  If  the 
number  of  such  elements  is  indefinitely  in- 
creased, the  line  joining  their  centroids 
approaches  as  a  limit  the  arc  AB\  And 
since  the  areas  of  the  elements  are  propor- 
tional to  the  lengths  of  the  corresponding 
portions  of  A'£',  the  centroid  of  the  total 
area  is  the  same  as  that  of  the  arc  A'B'. 
This  point  may  be  found  by  the  method  described  in  Art.  183. 

185.  Graphic  Determination  of  Areas.  —  Let  there  be  given 
any  number  of  geometrical  figures,  and  let  it  be  required  to 
determine  the  relative  magnitudes  of  their  areas. 

If  a  number  of  rectangles  can  be  found,  of  areas  equal 
respectively  to  the  areas  of  the  given  figures  and  having  one 
common  side,  then  the  remaining  sides  will  be  proportional  to 
the  areas  of  the  given  figures. 


M  N 


CEXTROIDS    OF    LINES    AND    OF   AREAS. 


157 


An  important  case  is  that  in  which  the  given  figures  are 
such  that  the  area  of  each  is  equal  to  the  product  of  two  known 
lines.  In  this  case  a  series  of  equivalent  rectangles  having  one 
common  side  can  be  found  by  the  following  construction. 

Let  ABC  (Fig.  63)  be  a  triangle,  and  let  it  be  required  to 
determine  an  equivalent 
rectangle  having  a  side  of 
given  length  as  LM.  Let 
b  and  h  be  the  base  and 
altitude  of  the  triangle. 

TVTn  lc  £*  T  J\/~ -• •-  h  ^\  n  c\  T  J~^ J?  ^-£*  Oo 

and  draw  the  semicircumference  PRN.     Draw  the  ordinate  LR 
perpendicular  to  PJV]  then 

LR2  =  PLx  LN=  I  bh  =  area  ABC. 
Draw  MR,  and  perpendicular  to  it  draw  RQ.     Then 

ABC. 


Hence  LQ  is  the  required  length. 

If  the  given  figure  is  a  parallelogram,  LN  and  LP  may  be 
its  base  and  altitude.  If  it  be  a  circular  sector,  LN  and  LP 
may  be  the  length  of  the  arc  and  half  the  radius. 

1 86.  Centroids  of  Partial  Areas.  —  It  may  be  required  to 
find  the  centroid  of  the  part  remaining  after  deducting  known 
parts  from  a  given  area.  For  this  case  the  construction  of 
Art.  182  needs  modification.  In  the  case  there  considered  we 
regarded  the  partial  areas  and  the  total  area  as  proportional  to 
parallel  forces  acting  at  their  respective  centers  of  gravity.  In 
this  case  we  may  also  represent  the  total  area,  the  portions 
deducted  from  it,  and  the  remaining  portion  as  forces  acting  at 
the  respective  centers  of  gravity ;  but  the  forces  corresponding 
to  the  areas  deducted  must  be  taken  as  acting  in  the  opposite 
direction  to  that  assumed  for  the  forces  representing  the  total 
area  and  the  area  remaining. 

Thus,  to  find  the  centroid  of  the  area  remaining  after  deduct- 


158 


GRAPHIC    STATICS. 


ing  from  the  circle  ABD  (Fig.  64)  a  smaller  circle  EFH  and  a 
sector  OAB,  we  may  proceed  as  follows  :  Find  the  centroid  of 
three  parallel  forces  proportional  to  the 
areas  ABD,  EFH,  and  OAB,  and  applied 
at  their  respective  centroids  O,  C,  and  G\ 
but  the  last  two  must  be  taken  as  acting 
in  the  direction  opposite  to  that  of  the  first. 
With  this  understanding,  the  force  and 
funicular  polygons  may  be  employed  as  in 
Art.  172. 

187.  Moments  of  Areas.  — The  moment  of  an  area  about  any 
line  in  its  plane  may  be  determined  from  the  funicular  polygon 
employed  in  finding  its  center  of  gravity.  Let  the  parallel 
forces  applied  at  the  centroids  of  the  partial  areas  be  assumed 
to  act  parallel  to  the  axis  of  moments.  Then  the  distance  inter- 
cepted on  the  axis  by  the  extreme  lines  of  the  funicular  poly- 
gon, multiplied  by  the  pole  distance,  is  equal  to  the  moment  of 
the  total  area  about  the  axis.  For,  by  Art.  56,  this  construc- 
tion gives  the  moment  of  the  resultant  force  about  any  point 
in  the  given  axis ;  and  this  is  equal  to  the  moment  of  the  result- 
ant area  about  the  axis  by  definition. 

A  similar  rule  gives  the  moments  of  the  partial  areas. 


CHAPTER    IX.       MOMENTS    OF    INERTIA. 

§   i.    Moments  of  Inertia  of  Forces. 

1 88.  Definitions. — The  moment  of  inertia  of  a  body  with 
respect  to  an  axis  is  the  sum  of  the  products  obtained  by  mul- 
tiplying the  mass  of  every  elementary  portion  of  the  body  by 
the  square  of  its  distance  from  the  given  axis. 

The  moment  of  inertia  of  an  area  with  respect  to  an  axis  is 
the  sum  of  the  products  obtained  by  multiplying  each  element- 
ary area  by  the  square  of  its  distance  from  the  axis. 

The  moment  of  inertia  of  a  line  may  be  similarly  defined, 
using  elements  of  length  instead  of  elements  of  mass  or  area. 

The  moment  of  inertia  of  a  force  with  respect  to  any  axis  is 
the  product  of  the  magnitude  of  the  force  into  the  square  of  the 
distance  of  its  point  of  application  from  the  axis.  The  sum  of 
such  products  for  any  system  of  parallel  forces  is  the  moment 
of  inertia  of  the  system  with  respect  to  the  given  axis. 

[NOTE.  —  The  term  moment  of  inertia  had  reference  originally  to  material  bodies, 
the  quantity  thus  designated  having  especial  significance  in  dynamical  problems  relat- 
ing to  the  rotation  of  rigid  bodies.  The  quantity  above  defined  as  the  moment  of 
inertia  of  an  area  is  of  frequent  occurrence  in  the  discussion  of  beams,  columns,  and 
shafts  in  the  mechanics  of  materials.  In  the  graphic  discussion  of  moments  of  iner- 
tia of  areas,  it  is  convenient  to  treat  areas  as  forces,  just  as  in  the  determination  of 
centers  of  gravity;  it  is  therefore  convenient  to  use  the  term  moment  of  inertia  of  a 
force  in  the  sense  above  defined.  It  is  only  in  the  case  of  masses  that  the  term 
moment  of  inertia  is  really  appropriate,  but  it  is  by  analogy  convenient  to  apply  it 
to  the  other  cases.] 

The  product  of  inertia  of  a  mass  with  respect  to  two  planes  is 
the  sum  of  the  products  obtained  by  multiplying  each  element- 
ary mass  by  the  product  of  its  distances  from  the  two  planes. 


I6o  GRAPHIC    STATICS. 

The  product  of  inertia  of  an  area,  a  line,  or  a  force  may  be 
defined  in  a  similar  manner. 

For  a  plane  area,  the  product  of  inertia  with  respect  to 
two  lines  in  its  plane  may  be  defined  as  the  sum  of  the 
products  obtained  by  multiplying  each  element  of  area  by  the 
product  of  its  distances  from  the  given  lines.  This  is  equiv- 
alent to  the  product  of  inertia  of  the  area  with  respect  to  two 
planes  perpendicular  to  the  area  and  containing  the  two  given 
lines. 

For  a  system  of  forces  with  points  of  application  in  the  same 
plane,  the  product  of  inertia  with  reference  to  two  axes  in  that 
plane  may  be  defined  as  the  sum  of  the  products  obtained  by 
multiplying  the  magnitude  of  each  force  by  the  product  of  the 
distances  of  its  point  of  application  from  the  given  axes.  It  is 
with  such  systems  and  with  plane  areas  that  the  following  pages 
chiefly  deal. 

The  radius  of  gyration  of  a  body  with  respect  to  an  axis  is  the 
distance  from  the  axis  of  a  point  at  which,  if  the  whole  mass  of 
the  body  were  concentrated,  its  moment  of  inertia  would  be 
unchanged.  The  square  of  the  radius  of  gyration  is  equal  to 
the  quotient  obtained  by  dividing  the  moment  of  inertia  of  the 
body  by  its  mass. 

The  radius  of  gyration  of  an  area  may  be  denned  in  a  similar 
manner. 

The  radius  of  gyration  of  a  system  of  parallel  forces  is 
the  distance  from  the  axis  of  the  point  at  which  a  force 
equal  in  magnitude  to  their  resultant  must  act  in  order  that 
its  moment  of  inertia  may  be  the  same  as  that  of  the  system. 
The  square  of  the  radius  of  gyration  may  be  found  by  dividing 
the  moment  of  inertia  of  the  system  by  the  resultant  of  the 
forces. 

189.  Algebraic  Expressions  for  Moment  and  Product  of  In- 
ertia.—  Moment  of  inertia.  —  Let  m^  ;«2,  etc.,  represent  ele- 
mentary masses  of  a  body,  and  rl9  ;-2,  etc.,  their  respective 


MOMENTS   OF    INERTIA   OF    FORCES.  J6I 

distances  from  an  axis  ;  then  the  moment  of  inertia  of  the  body 
with  respect  to  that  axis  is 


the  symbol  2  being  a  sign  of  summation,  and  the  second  mem- 
ber of  the  equation  being  merely  an  abbreviated  expression  for 
the  first. 

Product  of  inertia.  —  Let/i,  /2,  etc.,  denote  the  perpendicular 
distances  of  elements  m\,  m2,  etc.,  from  one  plane,  and  qlt  q^  etc., 
their  distances  from  another  ;  then  the  product  of  inertia  of  the 
body  with  respect  to  the  two  planes  is 

m\p\q\  4-  m*p?qz  H  ----  =  2mpq. 

Raditis  of  gyration.  —  With  the  same  notation,  if  k  denotes 
the  radius  of  gyration  of  the  body,  we  have 

(/;/!  +  mz-\  ----  )  k~  =  m  i  r?  +  m»r22  H  ----  . 
Or, 


7,2  _     \\         <>     H 


Here  2m  is  equal  to  the  whole  mass  of  the  body. 

Prodtict-radius.  —  Let  c  represent  a  quantity  defined  by  the 
equation. 

(mi  +  7#2  +  •  •  •)  <?  =  ^i\p\q\  +  Wa/2^2  +  "  '• 

Then  if  the  whole  mass  were  concentrated  at  the  same  distance 
c  from  two  axes,  its  product  of  inertia  with  respect  to  those 
axes  would  be  unchanged.  This  quantity  c  is  thus  seen  to  be 
analogous  to  the  radius  of  gyration.  It  may  be  called  the 
product-radius  of  the  body  with  respect  to  the  two  axes.  The 
value  of  c  is  always  given  by  the  equation 


Expressions  similar  to  those  just  given  apply  also  to  plane 
areas  and  to  systems  of  parallel  forces.  In  case  of  an  area, 
m\t  ;«2»  etc.,  denote  elements  of  area;  r^  r2,  etc.,  their  distances 
from  the  axis  of  inertia  ;  and  (plt  q^),  (/2,  ^2),  etc.,  their  dis- 


162  GRAPHIC    STATICS. 

tances  from  the  two  planes.  In  case  of  a  system  of  parallel 
forces  with  complanar  points  of  application,  m^  m2,  etc.,  must 
be  replaced  by  the  magnitudes  of  the  forces. 

190.  Determination  of  Moment  of  Inertia  of  a  System  of 
Parallel  Forces.  —  Let  the  points  of  application  of  the  forces 
be  in  the  same  plane,  which  also  contains  the  assumed  axis. 
We  shall  have  to  deal  only  with  systems  satisfying  these  condi- 
tions. By  the  definition,  the  moment  of  inertia  will  be  the 
same,  whatever  the  direction  of  the  forces.  If  we  take  the 
moment  of  any  force  (as  defined  in  Art.  175)  about  the  given 
axis,  and  suppose  a  force  equal  in  magnitude  to  this  moment 
to  act  at  the  point  of  application  of  the  original  force,  and  in  a 
direction  corresponding  to  the  sign  of  the  moment,  then  the 
moment  of  this  new  force  about  the  given  axis  is  equal  to 
the  moment  of  inertia  of  the  original  force.  If  this  be  done  for 
all  the  forces,  the  algebraic  sum  of  the  results  will  be  the 
required  moment  of  inertia  of  the  system. 

This  process  can  be  carried  out  graphically  by  methods 
already  described. 

Let  ab,  be,  cd,  de  (Fig.  65)  be  the  points  of  application  of 
four  parallel  forces,  and  let  the  axis  of  inertia  be  QR.  Suppose 
all  the  forces  to  act  in  lines  parallel  to  QR,  passing  through  the 
given  points  of  application.  Their  respective  moments  with 
reference  to  QR  may  now  be  found  by  the  method  of  Art.  55. 

Draw  the  force  polygon  ABCDE  and  choose  a  pole  O,  taking 
the  pole  distance  H  preferably  equal  to  AE  or  some  simple 
multiple  of  AE.  (In  Fig.  65  H  is  taken  equal  to  AE.)  Draw 
a  funicular  polygon  and  prolong  each  string  to  intersect  QR. 
Then  the  moment  of  any  force  with  respect  to  QR  is  the 
product  of  H  by  the  distance  intercepted  on  QR  by  the  two 
strings  corresponding  to  the  force  in  question.  Thus  the 
moment  of  AB  is  given  by  the  distance  A'BJ  (Fig.  65)  multi- 
plied by  H.  Also  the  successive  moments  of  BC,  CD,  DE  are 
represented  by  B*C\  C'D',  Df£',  each  multiplied  by  H.  It  is 


MOMENTS    OF    INERTIA    OF   FORCES. 


163 


seen  also  that  the  intercepts,  if  read  in  the  above  order,  give  a 
distinction  between  positive  and  negative  moments ;  upward 
distances  on  QR  denoting  in  this  case  positive  moments,  and 
downward  distances  negative  moments. 

We  have  now  to  find  the  sum  of  the  moments  of  a  second 
system  of  forces  acting  in  the  original  lines,  but  represented  in 


(-4) 


Fig.  65 

magnitude  and  direction  by  the  intercepts  just  found.  We 
may  take  as  the  force  polygon  for  the  second  construction  the 
line  A'B'C'D'E'  (Fig.  65),  and  choosing  any  pole  distance  H\ 
draw  a  second  funicular  polygon  and  find  the  distances  inter- 
cepted by  the  successive  strings  on  the  line  QR.  But  in  this 
case,  since  only  the  resultant  is  desired,  we  need  only  find  the 
intercept  AnE"  between  the  first  and  last  strings.  The  product 
of  this  intercept  by  H'  gives  the  sum  of  the  moments  of  A'BJ, 
B'C,  CD',  D!E'  with  respect  to  QR ;  and,  if  the  product  be 
multiplied  by  H  (since  A'B',  B'C',  etc.,  should  each  be  multi- 
plied by  H  in  order  to  represent  the  magnitudes  of  the  forces 
of  the  second  system),  the  result  will  be  the  required  moment  of 
inertia  of  the  given  system  of  forces  AB,  BC,  CD,  DE. 


1 64 


GRAPHIC    STATICS. 


It  should  be  noticed  that  in  Fig.  65  (A)  is  a  force  diagram, 
(B)  a  space  diagram  (Art.  u);  that  is,  every  line  in  (A)  repre- 
sents a  force,  while  every  line  in  (B)  represents  a  distance. 
Even  A'B',  B'C',  etc.,  though  used  as  forces,  are  actually 
merely  distances ;  and  the  moment  of  any  one  of  them  is  the 
product  of  a  length  by  a  length. 

191.  Radius  of   Gyration.  —  The  moment  of   inertia  of   the 
given  system  is  HxH'  xA"E".     If  H  has  been  taken  equal 
to  AB,  the  product  H'xA"E"  must  equal  the  square  of  the 
radius  of  gyration  of  the  system  with  respect   to   QR.      The 
length  of  the  radius  of  gyration  can  be  found  as  follows  (Fig. 
65) :  Draw  LN=H'  +  A"E",  and  make  LM=H'.     On  LN as 
a  diameter  draw  a  semicircle  LPN,  and  from  M  draw  a  line 
perpendicular  to  LN,  intersecting  the  semicircle  in  P.     Then 
MP  is  the  length  of  the  required  radius  of  gyration.     For  by 
elementary  geometry  we  have  PM2  =  LMxMN. 

If  H  is  taken  equal  to  n  x  AE,  the  moment  of  inertia  is  equal 
to  H'  xA"E"  x  ;/  x  AE,  and  the  square  of  the  radius  of  gyration 
is  equal  to  nH'xA"E".  Hence  in  Fig.  65  we  should  put 
either  LM=nH',  or  MN=nxA"E". 

192.  Central  Axis.  —  If  the  axis  with  reference  to  which  the 
moment  of  inertia  is  found  contains  the  centroid  of  the  given 
system  of  forces,  it  is  called  a  central  axis  of  the  system. 

In  many  cases  it  is  desired  to  find  the  moment  of  inertia  with 
respect  to  a  central  axis  whose  direction  is  known  while  the 
position  of  the  centroid  is  at  first  unknown.  It  is  to  be  noticed 
that  the  method  shown  in  Fig.  65  is  applicable  in  this  case  ; 
for  the  first  part  of  the  process  is  identical  with  that  employed 
in  finding  the  centroid  of  the  system.  If,  in  Fig.  65,  the 
strings  oat  oe,  of  the  first  funicular  polygon  be  prolonged  to 
intersect,  a  line  through  their  point  of  intersection,  parallel  to 
the  direction  of  the  forces,  will  contain  the  centroid  of  the 
system.  If  this  line  is  taken  as  the  inertia-axis,  the  points  A', 
B' ,  C',  D',  E'  are  the  points  in  which  this  axis  is  intersected  by 


MOMENTS    OF    INERTIA   OF   FORCES.  ^5 

the  strings  oa,  ob,  oc,  od,  oe.  No  further  modification  of  the 
process  is  necessary. 

193.  Moment  of  Inertia  Determined  from  Area  of  Funicular 
Polygon.  —  In  Fig.  65,  the  moments  of  the  given  forces  are 
represented  by  the  intercepts  A'B1,  B'C',  C'D't  D' E' ,  each 
multiplied  by  the  pole  distance  H.  The  moment  of  inertia  of 
any  force,  as  AB,  is  equal  to  the  moment  of  a  force  represented 
by  the  corresponding  intercept  as  A'B',  supposed  to  act  in  the 
line  ab.  Now,  by  definition  (Art.  175)  the  moment  about  the 
axis  QR  of  a*  force  equal  to  A'B'  acting  in  the  line  ab  is  equal 
to  d6uble  the  area  of  the  triangle  A'  i  B'  \  hence  the  moment  of 
the  force  HxA'B'  is  equal  to  double  the  area  of  that  triangle 
multiplied  by//.  Similarly,  the  moment  of  a  force  HxB'C', 
acting  in  the  line  be,  is  equal  to  double  the  area  of  the  triangle 
B'  2  C'  multiplied  by  H.  Applying  the  same  reasoning  to  each 
force,  we  see  that  the  sum  of  the  moments  of  the  assumed 
forces  (HxA'B1,  HxB'C',  etc.)  is  equal  to  2  H  times  the  sum 
of  the  areas  of  the  triangles  A1  i  B',  B'  2  C',  C'  3  D'y  D'  *£'. 
In  adding  these  triangles,  each  must  be  taken  with  its  proper 
sign,  corresponding  to  the  sign  of  the  moment  represented  by 
it.  Thus,  the  moments  of  A'B',  B'C',  and  D'E'-  all  have  the 
same  sign,  while  the  moment  of  C'D'  has  the  opposite  sign. 
Hence  we  must  have  for  the  sum  of  the  moments, 

area  ,4'  i  £'+area  B'  2  C'-area  C'  3  Z>'  +  area  D1  *£', 

which  is  equal  to  the  area  of  the  polygon  A'  i  2  34^'.  Hence 
this  area,  multiplied  by  2  H,  gives  the  moment  of  inertia  of  the 
required  system  of  forces. 

It  may  sometimes  be  convenient  to  apply  this  principle  in 
determining  moments  of  inertia,  the  area  being  determined  by 
use  of  a  planimeter,  or  by  any  other  convenient,  method.  It 
should  be  noticed  that  if  H  is  taken  equal  to  the  sum  of  the 
given  forces  (AE),  twice  the  area  of  the  funicular  polygon  is 
equal  to  the  square  of  the  radius  of  gyration.  If  H=  \  AE,  the 


1 66 


GRAPHIC    STATICS. 


square  of  the  radius  of  gyration  is  equal  to  the  area  of   the 
polygon. 

194.  Determination  of  Product  of  Inertia  of  Parallel  Forces. 
—  Assume  the  points  of  application  of  the  forces  to  be  in  the 
plane  containing  the  two  axes.  If  the  moment  of  any  force 
with  respect  to  one  axis  be  found,  and  a  force  equal  in  magni- 
tude to  this  moment  be  assumed  to  act  at  the  point  of  applica- 
tion of  the  original  force,  then  the  moment  of  this  new  force 
with  respect  to  the  second  axis  is  equal  to  the  product  of 
inertia  of  the  given  force  for  the  two  axes. 

Thus,  let  ab,  be,  cd,  de  (Fig.  66)  be  the  points  of  application 
of  four  parallel  forces,  and  let  their  product  of  inertia  with 
respect  to  the  axes  QR,  ST  be  required.  Draw  ABCDE,  the 


J^ 


force  polygon  for  the  given  forces,  assumed  to  act  parallel  to 
QR.  Choose  a  pole  O,  the  pole  distance  being  preferably 
taken  equal  to  AE,  or  some  simple  multiple  of  AE,  and  draw 
the  funicular  polygon  as  shown,  prolonging  the  strings  to  inter- 
sect QR  in  the  points  A',  B\  C\  D\  E'.  Now  assume  a  series 


MOMENTS    OF   INERTIA   OF   FORCES. 


I67 


of  forces  equal  to  A'B',  £'C',  etc.,  each  multiplied  by  H,  to  act 
at  the  points  ab,  be,  etc.,  and  determine  their  moments  with 
respect  to  the  axis  ST.  To  find  these  moments,  draw  lines 
through  ab,  be,  etc.,  parallel  to  ST,  and  draw  a  funicular  poly- 
gon for  the  assumed  forces  taken  to  act  in  these  lines.  The 
force  polygon  is  obtained  by  revolving  the  line  A'B'C'D'E'  until 
parallel  with  ST,  and  is  the  line  A\  B\  C\  D\  E\  in  the  figure. 
The  strings  o'a',  o'e'  of  the  second  funicular  polygon  intersect 
ST  in  the  points  A"  and  E".  Hence,  calling  H'  the  second 
pole  distance,  the  product  of  inertia  of  the  given  system  is 


195.  Product-Radius.  —  If  //"be  taken  equal  to  AE  (Fig.  66), 
H'xA"E"  is  equal  to  the    square  of  the  product-radius  (Art. 
189).     Hence  the  product-radius  can  be  found  by  a  construc- 
tion exactly  like  that  employed  in  finding  the  radius  of  gyra- 
tion.    Thus   (Fig.  66)  take   LM=H'  and  MN=A"E";  make 
LN  the  diameter  of  a  semicircle,  and  draw  from  Ma.  line  per- 
pendicular  to   LN,   intersecting   the    semicircle   in    P.     Then 
MP2=LMx  MN=  H'xA"E"\    hence    MP  =  c,    the    product- 
radius. 

196.  Relation  between  Moments  of  Inertia  for  Parallel  Axes. 

—  Proposition.  —  The  moment  of  inertia  of  a  system  of  par- 
allel forces  with  reference  to  any  axis  is  equal  to  its  moment  of 
inertia  with  respect  to  a  parallel  axis  through  the  centroid  of 
the  system  plus  the  moment  of  inertia  with  respect  to  the  given 
axis  of  the  resultant  applied  at  the  centroid  of  the  system. 

Let  PI,  P2,  etc.,  represent  the  forces  ;  x\,  x*  etc.,  the  dis- 
tances of  their  points  of  application  from  the  central  axis  ;  and 
a  the  distance  of  this  central  axis  from  the  given  axis.  Calling 
the  required  moment  of  inertia  A,  and  the  moment  of  inertia 
with  respect  to  the  axis  through  the  .center  of  gravity  A',  we 
have 

A=Pl 


168  GRAPHIC    STATICS. 

Now  P&  -f  P2xz  H  —  is  the  algebraic  sum  of  the  moments  of 
the  given  forces  with  respect  to  the  axis  through  their  centroid, 
and  is  equal  to  zero  ;  and  P^f  -f  P2x2  -  -\  —  =  A'.  Hence 

A=A'  +  (Pl  +  P,+  ...)a\ 
which  proves  the  proposition. 

Radii  of  gyration.  —  Let  k  =  radius  of  gyration  of  the  sys- 
tem with  respect  to  the  given  axis,  and  k'  the  radius  of  gyra- 
tion with  respect  to  the  central  axis,  and  we  have 


Hence  the  equation  above  deduced  may  be  reduced  to  the  form 


197.  Products  of  Inertia  with  Respect  to  Parallel  Axes.  - 
Proposition.  —  The  product  of  inertia  of  a  system  of  parallel 
forces  with  reference  to  any  two  axes  is  equal  to  the  product 
of  inertia  with  reference  to  a  pair  of  central  axes  parallel  to  the 
given  axes,  plus  the  product  of  inertia  of  the  resultant  (acting 
at  the  centroid)  with  reference  to  the  given  axes. 

Let  Pv  P2,  etc.,  be  the  magnitudes  of  the  given  forces  ;  (/,,  <?i), 
(/2,  ^2),  etc.,  the  distances  of  their  points  of  application  from 
the  central  axes  parallel  to  the  two  given  axes  ;  (a,  b)  the  dis- 
tances of  the  centroid  of  the  system  from  the  given  axes.  Let 
A  and  Af  be  the  products  of  inertia  of  the  system  with  respect 
to  the  given  axes  and  the  parallel  central  axes  respectively. 
Then 


A  =Pi(j>l 

-  (  /Vtfi  +  /Vtf*  +•••)+  a  (/Vi  +  P*q*  +  •  '  •) 


But  Pigi+Ptf»+--=o;  and  -Pi/i  +  A/^H  —  =o;  since  each  of 
these  expressions  is  the  sum  of  the  moments  of  the  given  forces 


MOMENTS    OF    INERTIA   OF   PLANE   AREAS  169 

with   respect   to   an   axis   through  the  centroicl   of  the   system. 
Hence, 


which  proves  the  proposition. 

If  A  =  (Pl  +  P,+  --)randA>=(Pl  +  P,+  -.-y2,  we  have 


From  the  above  proposition  it  follows  that  if  the  axes  have 
such  directions  that  the  product  of  inertia  with  reference  to  the 
central  axes  is  zero,  the  product  of  inertia  with  reference  to  the 
given  axes  is  the  same  as  if  the  forces  all  acted  at  the  centroicl. 
When  this  condition  is  known  to  be  satisfied,  then  for  the  pur- 
pose of  finding  the  product  of  inertia  the  system  of  forces  may 
be  replaced  by  their  resultant. 

It  follows  also,  in  the  case  when  the  product  of  inertia  for  the 
central  axes  is  zero,  that  if  one  of  the  given  axes  coincides  with 
the  parallel  central  axis,  the  product  of  inertia  for  the  given 
axes  is  zero  ;  for  in  this  case  either  a  or  b  is  zero,  and  hence 
ab(Pi  +  Pz-\  ----  )  is  zero.  Therefore, 

If  the  product  of  inertia  of  a  system  is  zero  for  two  axes,  A' 
and  A",  one  of  which  (as  A')  contains  the  centroid  of  a  system, 
then  the  product  of  inertia  is  also  zero  for  A1  and  any  axis 
parallel  to  A". 

§   2.    Moments  of  Inertia  of  Plane  Areas. 

198.  Elementary  Areas  Treated  as  Forces.  —  If  any  area  be 
divided  into  small  elements,  and  a  force  be  applied  at  the 
centroid  of  each  element  numerically  equal  to  its  area,  the 
moment  of  inertia  of  this  system  of  forces  will  be  approximately 
equal  to  that  of  the  given  area.  The  approximation  will  be 
closer  the  smaller  the  elementary  areas  are  taken.  If  the  ele- 
ments be  made  smaller  and  smaller,  so  that  the  area  of  each 
approaches  zero  as  a  limit,  the  moment  of  inertia  of  the  sup- 
posed system  of  forces  approaches  as  a  limit  the  true  value  of 
the  moment  of  inertia  of  the  given  area. 


GRAPHIC   STATICS. 

It  is  seen,  then,  that  most  of  the  general  principles  which 
have  been  stated  regarding  moments  of  inertia  of  systems  of 
forces  are  equally  applicable  to  moments  of  inertia  of  areas. 
The  practical  application  of  these  principles,  however,  and 
especially  the  graphic  constructions  based  upon  them,  are  less 
simple  in  the  case  of  areas  than  of  systems  of  forces  such  as 
those  already  treated.  The  reason  for  this  is  that  the  system 
of  forces  which  may  be  conceived  to  replace  the  elements  of 
area  consists  of  an  infinite  number  of  infinitely  small  forces, 
with  which  the  graphic  methods  thus  far  discussed  cannot 
readily  deal.  Problems  of  this  class  are  most  easily  treated  by 
means  of  the  integral  calculus,  especially  when  the  areas  dealt 
with  are  in  the  form  of  geometrical  figures.  It  is  possible, 
however,  by  graphic  methods  to  determine  approximately  the 
moment  of  inertia  of  any  plane  area  ;  and  in  many  cases  exact 
graphic  solutions  of  such  problems  are  not  difficult.  The  proof 
of  these  methods  is  often  most  easily  affected  algebraically. 

199.  Moments  of  Inertia  of  Geometrical  Figures.  —  The  appli- 
cation of  the  integral  calculus  to  the  determination  of  moments 
of  inertia  will  not  be  here  treated.  But  the  values  of  the 
moments  of  inertia  of  some  of  the  common  geometrical  figures 
are  of  such  frequent  use  that  the  more  important  of  them  will 
be  given  for  future  reference.  The  moment  of  inertia  is  in  each 
case  taken  with  respect  to  a  central  axis,  and  will  be  repre- 
sented by  /,  while  the  radius  of  gyration  will  be  called  k. 

Rectangle.  —  Let  b  and  d  be  the  sides,  the  axis  being  parallel 
to  the  side  b.  Then 

A_^3.      /a-*'2 
=  12  '          ~7J 

Triangle.  —  Let  b  and  d  be  the  base  and  altitude.  Then  for 
an  axis  parallel  to  the  base, 


36' 


MOMENTS    OF   INERTIA   OF   PLANE   AREAS. 


I/I 


For  an  axis   through  the   vertex,  bisecting  the  base,   k^  —  —  , 

24 
where  b*  is  the  projection  of  the  base  on  a  line  perpendicular  to 

the  axis. 

Circle.  —  Let  d  be  the  diameter.     Then 

r^4.        p_d? 

—  ~7-       »        K   ~       s-' 

64  16 

For  a  central  axis  perpendicular  to  the  plane  of  the  circle, 


32  '  8 

Ellipse.  —  Let  a  and  b  be  the  semi-axes.     Then  for  an  axis 
parallel  to  a, 


4 
For  an  axis  parallel  to  b, 


/—  TTt       .         A2  _ 

4  4 

For  a  central  axis  perpendicular  to  the  plane  of  the  ellipse, 


Graphic  construction  for  radius  of  gyration.  —  Whenever  £2 
can  be  expressed  as  the  product  of  two  known  factors,  the  value 
of  k  can  be  found  by  the  construction  already  used  in  Art.  191. 

/2 

Thus,  in  case  of   a  rectangle,  for  which  k*=  —  ,  we  may  put 

p  =  d    d     Then  if  in  p.      6     we  take  LM=d  MN=d,  the 

34  43 

construction  there  shown  will  give  MP  as  the  value  of  k.     For 

the   triangle,    the   axis   being   parallel    to   the   base,    we   have 
£2  =  -  •  -,  and  the   same  construction  is  applicable.     For  the 

36  v  v 

axis  through  the  vertex  bisecting  the  base,  kz  =  —  -  —  • 

4      6 


1/2 


GRAPHIC   STATICS. 


200.  Product  of  Inertia.  —  General  principles.  —  Products  of 
inertia  of  areas  are  determined  by  means  of  the  integral  calculus 
in  a  manner  similar  to  that  employed  for  moments  of  inertia. 
The   following    fundamental    principles    regarding   products  of 
inertia  of  geometrical  figures  will  be  found  useful : 

(1)  With  reference  to  two  rectangular  axes,  one  of  which  is 
an  axis  of  symmetry  (Art.  179),  the  product  of  inertia  is  zero. 
For  it  is  manifest   that   the  products  of   inertia  of  two  equal 
elements,   symmetrically  placed  with  reference  to  one  of   the 
axes,  are  numerically  equal  but  of  opposite  sign.      Hence,  if 
the  whole  area  can  be  made  up  of  such  pairs  of  elements,  the 
total  product  of  inertia  is  zero. 

(2)  If  the  two  axes  are  not  rectangular,  but  the  area  can  be 
divided  into  elements  such  that  for  every  element  whose  dis- 
tances from  the  axes  are  /,  q,  there  is  an  equal  element  whose 
distances  are  /,  —  q,  or   — /,  q,  the  product  of  inertia  is  zero. 
This  includes  the  preceding  as  a  special  case. 

20 1.  Products  of  Inertia  of  Geometrical  Figures.  —  In  each  of 
the  following  cases  the  product  of  inertia  is  zero : 

A  triangle,  one  axis  containing  the  vertex  and  the  middle 
point  of  the  base,  the  other  being  any  line  parallel  to  the  base. 

A  parallelogram,  the  axes  being  parallel  to  the  sides  and  one 
axis  being  central.  This  includes  the  rectangle  as  a  special 
case. 

An  ellipse,  the  axes  being  parallel  to  a  pair  of  conjugate 
diameters,  and  one  axis  being  central.  This  includes,  as  a 
special  case,  that  in  which  one  axis  is  a  principal  diameter  and 
the  other  is  any  line  perpendicular  to  it  ;  and  under  this  case 
falls  also  the  circle. 

202.  Approximate  Method  for  Finding  Moment  of  Inertia  of 
Any  Area.  — To  apply  the  method  of  Art.  190  to  the  determina- 
tion of  the  moment  of  inertia  of  a  plane  area,  we  should  strictly 
need  to  replace  the  area  by  an  infinite  number  of  parallel  forces, 
proportional  to  the  infinitesimal  elements  of  the  given  area,  and 


MOMENTS    OF   INERTIA   OF   PLANE   AREAS. 


1/3 


with  points  of  application  in  these  elements.  If,  instead,  we 
divide  the  given  figure  into  finite  portions  whose  several  areas 
are  known,  and  assume  forces  proportional  to  those  areas  to 
act  at  their  centroids,  we  may  get  an  approximate  value  for  the 
moment  of  inertia,  which  will  be  more  nearly  correct  the  smaller 
the  elements.  This  will  be  illustrated  by  the  area  shown  in 
Fig.  67. 

Let  QR  be  the  axis  with  reference  to  which  the  moment  of 
inertia  is  to  be  found,  in  this   case  taken  as  a  central  axis. 


Fig.  or 


Divide  the  figure  into  four  rectangular  areas  as  shown,  and 
assume  forces  numerically  equal  to  these  areas  to  act  at  their 
centroids  parallel  to  QR.  The  force  polygon  is  ABODE. 
Draw  the  funicular  polygon  corresponding  to  a  pole  O,  and  let 
the  successive  strings  intersect  QR  in  A',  B1 ,  C',  Df,  £f.  Take 
this  as  a  new  force  polygon,  and  with  any  convenient  pole 
distance  draw  a  second  funicular  polygon,  using  the  same  lines 
of  action.  Let  the  first  and  last  strings  intersect  QR  in  A" 
and  E".  Then  A" E"  multiplied  by  the  product  of  the  two 
pole  distances  gives  the  moment  of  inertia  of  the  four  assumed 
forces,  and  approximately  the  moment  of  inertia  of  the  given 


174 


GRAPHIC   STATICS. 


figure.  If  the  first  pole  distance  is  taken  equal  to  AE  (as  is 
the  case  in  Fig.  67),  the  radius  of  gyration  may  be  found  by  the 
construction  of  Art.  191.  Thus  in  Fig.  67,  MP  is  the  radius 
of  gyration  as  determined  by  this  method. 

A  more  accurate  result  may  be  reached  by  dividing  the  area 
into  narrower  strips  by  lines  parallel  to  QR  ;  since  the  narrower 
such  a  strip  is,  the  more  nearly  will  the  distance  of  each  small 
element  from  the  axis  coincide  with  that  of  the  centroid  of  the 
strip.  If  the  partial  areas  are  taken  as  narrow  strips  of  equal 
width,  the  forces  may  be  taken  proportional  to  the  average 
lengths  of  the  several  strips. 

203.  Accurate  Methods.  —  If  the  given  figure  can  be  divided 
into  parts,  such  that  the  area  of  each  is  known,  and  also  its 
radius  of  gyration  with  respect  to  its  central  axis  parallel  to  the 
given  axis,  the  above  method  may  be  so  modified  as  to  give  an 
accurate  result.  Two  methods  will  be  noticed. 

(1)  When  the  axis  is  known  at  the  start.  —  Let  the  line  of 
action  of  the  force  representing  any  partial  area  be  taken  at  a 
distance  from  the  given  axis  equal  to  the  radius  of  gyration  of 
that  area  with  reference  to  the  axis.     If  this  is   done,   it  is 
evident  that  the  moment  of  inertia  of  the  system  of  forces  is 
identical  with  that  of  the  given  area.     When  the  axis  is  known, 
the  position  of  the  line  of  action  for  any  force  may  be  found  as 

follows  :  Let  QR  (Fig.  68)  be  the  given  axis, 
and  Q'R'  a  parallel  axis  through  the  centroid 
of  any  partial  area.  Draw  KL  perpendicu- 
lar to  QR,  and  lay  off  LM  equal  to  the 
radius  of  gyration  of  the  partial  area  with 
respect  to  Q'R'.  Then  KM  is  the  length 

of  the  radius  of  gyration  with  respect  to  QR.     (Art.  196.)     Take 

KN equal  to  KM  and  draw  Q"R"  through  N  parallel  to  Q'R' ; 

then   Q"R"  is  to  be  taken  as  the  line  of  action  of  the  force 

representing  the  partial  area  in  question. 

(2)  When  the  axis  is  at  first  unknown.  —  The  method  to  be 


MOMENTS   OF    INERTIA    OF   PLANE   AREAS. 


employed  in  this  case  is  to  let  the  force  representing  any  partial 
area  act  in  a  line  through  the  centroid  of  that  area  ;  and  then 
assume  the  force  representing  its  moment  to  act  in  such  a  line 
that  the  moment  of  this  second  force  shall  be  numerically 
equal  to  the  moment  of  inertia  of  the  partial  area.  This  line 
may  be  found  as  follows  :  Let  k  represent  the  radius  of  gyra- 
tion of  the  partial  area  with  respect  to  its  central  axis  parallel 
to  the  given  axis,  and  a  the  distance  between  the  two  axes. 
Then  the  moment  of  inertia  of  the  partial  area  with  respect 
to  the  given  axis  is  A  (<?2-f /£2),  if  A  represents  the  area.  But 

f       $\ 
A  (a2  +  &)  =  Aa{  a -\ — ).     Hence,  if  a  force  numerically  equal  to 

V       a) 

A  is  assumed  to  act  with  an  arm  a,  then  a  force  equal  to  its 

k2 
moment  Aa  must  act  with  an  arm  a-\-~  in  order  that  its  moment 

a 
may  equal  A  (<?2  +  £2). 

£2 

The  distance  a-\ —  can  be  found  by  a  simple  construction. 
a 

Let  QR  and  Q 'R'  (Fig.  69)  be  the  given  axis  and  the  central 
axis  respectively.  Draw  KL 
perpendicular  to  QR  and  lay  off 
L M  equal  to  k.  From  M  draw 
a  line  perpendicular  to  KM,  in- 
tersecting KL  produced  at  N. 

Then  KN=a  +  k--     For>  in  the 
a 

similar  triangles  KNM,  KML,  we  have 


.  09 


KN 


KM     KL 


But  KL=a,  and 


or  KN= 


hence 


KL 


This  second  method  is  more  useful  than  the  first,  because  in 
applying  it  the  first  funicular  polygon  can  be  drawn  before  the 
position  of  the  inertia-axis  is  known.  Thus,  a  very  common 


GRAPHIC    STATICS. 

case  is  that  in  which  the  moment  of  inertia  of  an  area  is  to  be 
found  for  a  central  axis,  whose  direction  is  known,  while  at  the 
outset  its  position  is  unknown  because  the  centroid  of  the  area 
is  unknown.  If  the  second  method  be  employed,  the  first 
funicular  polygon  can  be  drawn  at  once,  and  serves  to  locate 
the  required  central  axis,  as  well  as  to  determine  the  moments 
of  the  first  set  of  forces  as  soon  as  the  axis  is  known.  The 
central  axis  and  the  moment  of  inertia  with  respect  to  it  are 
thus  determined  by  a  single  construction. 

Example. — The  method  last  described  is  illustrated  in  Fig. 
70.  The  area  shown  consists  of  two  rectangles,  the  centroids 
of  which  are  marked  ab  and  be.  The  moment  of  inertia  is  to 
be  found  for  a  central  axis  parallel  to  the  longer  side  of  the 
rectangle  ab. 


Fig.  ro 


We  draw  through  ab  and  be  lines  parallel  to  the  assumed 
direction  of  the  axis,  and  take  these  for  the  lines  of  action  of 
forces  AB,  BC,  proportional  to  the  areas  of  the  two  rectangles. 
ABC  is  the  force  polygon,  and  the  pole  distance  is  taken  equal 


MOMENTS   OF    INERTIA   OF   PLANE   AREAS. 


1/7 


to  AC.  The  intersection  of  the  strings  oa,  oc,  determines  a 
point  of  the  required  central  axis  ac.  The  moments  of  the 
given  forces  are  proportional  to  A'B',  B'C.  We  now  have  to 
find  the  lines  of  action  for  the  forces  AB',  B'C,  in  accordance 
with  the  method  above  described. 

Take  any  line  perpendicular  to  the  axis  ac,  as  VQ,  the  side 
of  one  of  the  rectangles.  From  R,  the  point  in  which  this  line 
intersects  the  vertical  line  through  be,  lay  off  RT  equal  to  the 
central  radius  of  gyration  of  the  rectangle  whose  centroid  is 
be.  To  find  this  central  radius  of  gyration,  we  know  that  its 

value  is  ^ —  (Art.  199),  where  d  is  the  length  of  the  side  of 
the  rectangle  perpendicular  to  the  axis.  Hence  we  take 

QR  =  -,  RS=~,  and    make  QS  the  diameter  of  a  semicircle, 
3  4 

intersecting  the  vertical  line  be  in  T;  then  RT  is  the  required 
radius  of  gyration  of  the  rectangle  with  respect  to  a  central 
axis.  Now  draw  from  T  a  line  perpendicular  to  FT)  intersect- 
ing VQ  in  U;  then  the  line  b'c'  drawn  through  U  parallel  to  the 
given  axis  is  the  line  of  action  of  the  force  B'C'. 

By  a  similar  construction  applied  to  the  other  rectangle,  a't>' 
is  located  as  the  line  of  action  of  the  force  A'B'.  The  second 
funicular  polygon  is  now  drawn,  and  the  points  A11,  C"  are 
found  by  the  intersection  of  the  strings  o'a',  o'c1  with  the  axis. 
Hence  the  moment  of  inertia  of  the  given  area  is  equal  to 
A"C"xHxff'.  In  the  figure  H  is  made  equal  to  AC,  hence 
the  radius  of  gyration  can  be  determined  by  the  usual  con- 
struction, and  its  length  is  found  to  be  MP. 

204.  Moment  of  Inertia  of  Area  Determined  from  Area  of 
Funicular  Polygon.  —  The  method  given  in  Art.  193  for  finding 
the  moment  of  inertia  of  a  system  of  forces  by  means  of  the 
area  of  the  funicular  polygon  may  be  applied  with  approximate 
results  to  the  case  of  a  plane  figure.  If  the  forces  are  taken 
as  acting  at  the  centroids  of  the  areas  they  represent,  then  to 


I7g  GRAPHIC   STATICS. 

get  good  results  these  partial  areas  should  be  taken  as  narrow 
strips  between  lines  parallel  to  the  axis. 

If  the  lines  of  action  are  determined  as  in  the  first  method 
of  the  preceding  article,  then  the  area  enclosed  by  the  funic- 
ular polygon  and  the  axis  represents  accurately  the  required 
moment  of  inertia. 

205.  Product  of  Inertia  Determined  Graphically.  —  To  deter- 
mine the  product  of  inertia  of  any  area,  let  it  be  divided  into 
small  known  parts,  and  let  parallel  forces  numerically  equal  to 
the  partial  areas  be  assumed  to  act  at  the  centroids  of  these 
parts.  The  product  of  inertia  of  these  forces  may  then  be 
found  as  in  Art.  194,  and  its  value  will  represent  approximately 
the  product  of  inertia  of  the  given  area. 

If  the  partial  areas  can  be  so  taken  that  the  product  of 
inertia  of  each  with  reference  to  axes  through  its  centroid  par- 
allel to  the  given  axes  is  zero,  the  method  here  given  is  exact. 
(Art.  197.) 

If  the  partial  areas  are  taken  as  narrow  strips  parallel  to  one 
of  the  axes,  the  condition  just  mentioned  will  be  nearly  fulfilled  ; 
for  the  product  of  inertia  of  each  strip  for  a  pair  of  axes 
through  its  centroid,  one  of  which  is  parallel  to  its  length,  will 
be  very  small. 


CHAPTER    X.      CURVES    OF    INERTIA. 

§    I.    General  Principles. 

206.  Relation  between  Moments  of  Inertia  for  Different  Axes 
through  the  Same  Point.  — The  moment  of  inertia  with  respect 
to  any  axis  through  a  given  point  can  be  expressed  in  terms  of 
the  moments  and  product  of  inertia  for  any  two  axes  through 
that  point.  It  is  necessary  here  to  use  algebraic  methods,  but 
the  results  reached  form  the  basis  of  graphic  constructions. 

Let  OX,  <9F(Fig.  71)  be  the  two  given  axes;  6,  the  angle 
included  between  them  ;  Plt  P2)  etc.,  the  forces  of  the  system  ;  x1 1 
y',  the  coordinates  of  the  point 
of  application  of  any  force  P, 
referred  to  the  axes  OX,  OY; 
p,  q,  the  perpendicular  distances 
of  the  same  point  from  O  Y  and 
OX  respectively,  so  that 

p  =  x'  sin  6,    q=y'  sin  0. 
Let  a\  b',  and  c'  be  quantities  denned  by  the  equations 


a '  = 


179 


l8o  GRAPHIC    STATICS. 

Then 


and  #'  sin  0,  £'  sin  0,  r'  sin  0  are  respectively  the  radius  of 
gyration  with  respect  to  O  Y,  the  radius  of  gyration  with  respect 
to  OX,  and  the  product-radius  (Art.  189)  with  respect  to  OX 
and  OY. 

The  moment  of  inertia  (/)  and  radius  of  gyration  (k)  of  the 
system  for  the  axis  OM,  making  an  angle  <f>  with  OX,  may  now 
be  computed  as  follows  : 

Let  s  =  perpendicular  distance  of  the  point  of  application  of 
any  force  P  from  OM.  Then  from  the  geometry  of  the  figure 
it  is  seen  that 

s=y'  sin  (0  —  <f>)—^f  sin  <£. 
Hence 


sn       -<-2  sn      -      sn 


or, 

I=b^P  •  sin2  (0-$)-2c'*2P  •  sin  (0-0)  sin  <f> 


sn 


the  factors  involving  0  and  <£  being  constant  for  all  terms  of 
the  summation.     Hence 


=  --  =  b'*  sin2  (<9  -(/>)-  2  ^'2  sin  ((9-0)  sin  <£+V2  sin2  0    .   .    (i) 


From  these  equations  /and  £  may  be  computed  if  a',  b',  and 
c'  are  known  ;  that  is,  if  the  moments  and  product  of  inertia  for 
the  two  axes  (Wand  OYare  known. 

Special  case.  —  If  #  =  90°,  the  equation  (i)  becomes 


GENERAL   PRINCIPLES.  !8r 

207.  Products  of  Inertia  for  Different  Axes  through  the  Ori- 
gin. —  The  product  of  inertia  with  respect  to  OM  and  OX  may 
be  found  as  follows  : 

Let    A  —  the  required  product  of  inertia  ;  then 

^  =  2/V  =  2/y  sin  e  [/  sin  (0-0)-*'  sin  </>] 
=  2P  [/2sin0  sin  (0  -<£)-*'/  sin  6  sin  </>] 
=  [b'*  sin  0  sin  (<9-<£)-r'2  sin  <9  sin  0]  2/> 
Let  k  =  product-radius  for  axes  OM  and  OX\  then 

//2  =  A  =  ^2  sin  0  sin  ((9-0)-^2  sin  0  sin  0. 

Special  case.  —  The  axis  6W  may  be  so  chosen  that  A=o. 
This  will  be  the  case  if 

b'2  sin  ((9  -</>)=  ^'2  sin<£. 

208.  Inertia  Curve.  —  If  on  <9J/  (Fig.  71)  a  point  J/  be  taken 
such  that  the  length  O  M  depends  in  some  given  way  upon  the 
value  of  k,  and  if  similar  points  be  located  for  all  possible  direc- 
tions of  OM,  the  locus  of  such  points  will  be  a  curve  which  is 
called  a  curve  of  inertia  of  the  system  for  the  center  O. 

The  form  of  the  curve  will  depend  upon  the  assumed  law 
connecting  OM  with  k. 

209.  Ellipse  or  Hyperbola  of  Inertia.  —  The  simplest  curve  is 
obtained   by   assuming  OM  to  be  inversely  proportional  to  k, 
Let   OM=r,  and    take   r*  =  —  ^  —  >    where    <^2    is    a    positive 

K 

quantity,  so  that  d  always  represents  a  real  length,  positive  or 
negative. 

Equation  (i)  of  Art.   206  then  becomes 


which  is  the  polar  equation  of  the  inertia-curve,  r  and  <f>  being 
the  variable  coordinates.  Let  x,  y  be  coordinates  of  the  point 
M  referred  to  the  axes  OX,  O  Y.  Then 

r    _         x  y 

sin  0     sin  (#  —  <£)     sin  0 


1 82  GRAPHIC    STATICS, 

whence  — sin  (   —  q>)  _* 


sin2<9 
2  sin  (0- 


sin20 
and  the  equation  becomes 

The  form  of  this  equation  shows  that  it  represents  a  conic 
section  whose  center  is  at  the  origin  of  coordinates  O.  This 
conic  may  be  either  an  ellipse  or  a  hyperbola. 

The  equation  will  be  discussed  more  fully  in  a  later  article. 
One  fact  may,  however,  be  here  noticed. 

If  the  moment  of  inertia  /  is  positive  for  all  positions  of  the 
axis,  the  radius  of  gyration  k  will  be  real,  whatever  the  value 
of  <f>.  But  r,  the  radius  vector  of  the  curve,  will  be  real  when  k 
is  real ;  hence  in  this  case  the  curve  is  an  ellipse.  This  is 
always  the  case  if  the  given  forces  have  all  the  same  sign. 

If  the  forces  have  not  all  the  same  sign,  it  is  possible  that 
the  value  of  /  may  have  different  signs  for  different  directions 
of  the  axis.  If  this  is  so,  certain  values  of  <£  make  k  (and 
therefore  r)  imaginary.  In  this  case  the  curve  is  a  hyperbola. 

The  most  important  case  is  that  in  which  the  given  forces 
have  all  the  same  sign  which  may  be  taken  as  plus,  so  that  the 
moment  of  inertia  is  always  positive,  and  the  curve  an  ellipse ; 
and  to  this  case  the  discussion  will  be  confined. 

§  2.    Inertia-Ellipses  for  Systems  of  Forces. 

210.  Properties  of  the  Ellipse.  —  In  discussing  ellipses  of  in- 
ertia use  will  be  made  of  certain  general  properties  of  the 
ellipse,  which,  for  convenience  of  reference,  will  be  here  sum- 
marized. For  the  proof  of  the  propositions  stated  the  reader 
is  referred  to  works  on  the  conic  sections. 


INERTIA-ELLIPSES   FOR   SYSTEMS   OF   FORCES.         183 
(i)  The  equation 


represents    an   ellipse  if  EP—AC  is  negative;   a  hyperbola  if 
EP  —  AC  is  positive.     (Salmon's  Conic  Sections,  p.    140.)     The 

coordinate  axes  may  be  either  rectangular  or  oblique. 

• 

(2)  Two  diameters  of  an  ellipse  are  said  to  be  conjugate  to 
each  other  if  each  bisects  all  chords  parallel  to  the  other.  If 
the  axes  of  coordinates  coincide  with  a  pair  of  conjugate 
diameters,  the  lengths  of  which  are  2  a'  and  2  b',  the  equation 
of  the  curve  is 


A  particular  case  of  this  equation  is  that  in  which  the  coordi- 
nate axes  are  rectangular,  being  the  principal  axes  of  the  curve  ; 
in  which  case  we  may  write  a  and  b  instead  of  a'  and  b'. 

(3)  In  an  ellipse,  the  product  of  any  semi-diameter  and  the 
perpendicular  from  the  center  on  the  tangent  parallel  to  that 
semi-diameter  is  constant  and  equal  to  ab.     That  is,  if  r  is  any 
radius  vector  of  the  curve  drawn  from  the  center,  and  /  the 
length   of  the   perpendicular  from  the  center  to  the  parallel 
tangent,  we  have 

pr=ab 
where  a  and  b  are  the  principal  semi-axes  of  the  curve. 

(4)  Let  a'  and  b1  be  conjugate  semi-  diameters.     Then  each  is 
parallel  to  the  tangent  at  the  extremity  of  the  other.     Hence 
the  length  of  the  perpendicular  from  the  center  to  the  tangent 
parallel  to  a'  is  b'  sin  6,  where  0  is  the  angle  included  between 
a'  and  b'.     Therefore  from  the  preceding  paragraph, 

a'&'s'm  6  =  ab. 

(5)  An  ellipse  can  be  constructed,  when  a  pair  of  conjugate 
diameters  is  known,  as  follows  : 


1  84  GRAPHIC    STATICS. 

Let  AAf,  BB'  (Fig.  72),  be  the  conjugate  diameters,  O  being 
the  center  of  the  ellipse.  Complete  the  parallelogram  OBCA. 

Divide  OA  and  CA  into  parts 
proportional  to  each  other,  be- 
ginning at  O  and  C.  Through 
the  points  of  division  of  OA 
draw  lines  radiating  from  B1  ', 
and  through  the  points  of  divi- 
sion of  CA  draw  lines  radiating 
from  B.  The  points  of  inter- 
section of  the  corresponding  lines  in  the  two  sets  are  points  of 
the  ellipse.  In  a  similar  way,  the  other  three  quadrants  may 
be  drawn.  (The  location  of  one  point  is  shown  in  the  figure.) 

A  convenient  way  to  locate  the  corresponding  points  of 
division  on  OA  and  CA  is  to  cut  these  lines  by  lines  parallel  to 
the  diagonal  OC. 

211.  Discussion  of  Equation  of  Inertia-Curve.  —  We  will  now 
examine  the  equation  of  the  inertia-curve, 


with  reference  to  the  properties  of  the  ellipse  above  enumerated. 

(i)  If  dW2-V4  is  positive,  the  equation  denotes  an  ellipse. 
This  cannot  be  the  case  if  a'2  and  b'2  have  opposite  signs.  But 
from  the  definitions  of  af2  and  b^  (Art.  206)  it  is  seen  that  their 
signs  are  the  same  as  those  of  the  moments  of  inertia  for  Y  and 
X  axes  respectively.  Hence,  if  there  are  any  two  axes  through 
the  assumed  center  for  which  the  moments  of  inertia  have 
opposite  signs,  the  inertia-curve  is  a  hyperbola. 

If  the  moment  of  inertia  has  the  same  sign  for  all  axes 
through  the  assumed  center,  the  curve  is  an  ellipse.  For,  since 

by  choosing  the 


axes  so  that  the  product  of  inertia  with  respect  to  them  is 
zero  ;  and  if  c'  is  zero,  and  a'2  and  b™  have  the  same  sign,  the 
quantity  a'2b'2  —  c'*  is  positive. 


INERTIA-ELLIPSES   FOR   SYSTEMS   OF   FORCES.         185 

This  agrees  with  the  conclusion  stated  in  Art.  209. 
We  shall  here  deal  only  with  ellipses  of  inertia. 

(2)  If  <:f  =  o,  the  coordinate  axes  are  conjugate  axes  of   the 
curve.     But    the    condition   c'  =  o   means   that  the   product    of 
inertia  for  the  two  axes  is  zero.     Hence  any  two  axes  for  which 
the  product  of  inertia  is  zero  are  conjugate  axes  of  the  inertia- 
curve.     (This  is  true   whether   the   curve   is   an    ellipse   or   a 
hyperbola.) 

(3)  By  the  law  of  formation  of  ,,the  inertia-conic  (Art.  209), 
the  length  of  the  radius  vector  lying  in  any  line  is  inversely 
proportional  to  the  radius  of  gyration  with  respect  to  that  line. 
But  by  (3)  of  the  last  article,  the  perpendicular  from  the  center 
on  the  tangent  parallel  to  any  radius  vector  is  inversely  propor- 
tional to  the  length  of  that  radius  vector.     Hence  the  perpen- 
dicular distance  between  any  diameter  and  the  parallel  tangent 
is  directly  proportional  to  the  radius  of  gyration  with  respect  tc 
that    diameter.     The   curve    may  be  so   constructed   that    the 
length  of  this  perpendicular  is  equal  to  the  radius  of  gyration, 
as  follows  : 

From  Art.  209,  we  have 


r 

and  from  (3)'  and  (4)  of  the  last  article  we  have 

ab     a'b1  sin  6 

p  —  —  =  —          —  , 
r  r 

if  a'  and  b1  are  conjugate  semi-diameters.     Now  take 

d'2s'm  0  =  a&  =  a'6's'm0, 
or  d*=a'b\ 

and  we  have  k=p, 

and  the  equation  of  the  curve  becomes  (since  c'  =  o  when  the 
axes  are  conjugate) 


1  86  GRAPHIC    STATICS. 

If  the  equation  be  written  in  this  form,  a'  and  V  having  the 
meanings  assigned  in  Art.  206,  the  radius  of  gyration  about  any 
axis  through  the  center  of  the  ellipse  is  equal  to  the  perpendicular 
distance  between  the  axis  and  the  parallel  tangent  to  tJie  ellipse. 

Hereafter  we  shall  mean  by  inertia-ellipse  the  curve  obtained 
by  taking  d2  =  a'&'  as  above  described,  so  that  the  radius  of  gyra- 
tion for  any  axis  can  be  found  by  direct  measurement  when  a 
parallel  tangent  to  the  ellipse  is  known. 

212.  To  Determine  Tangents  to  the  Inertia-Ellipse  for  Any 
Center.  —  Let    the  radius   of   gyration    (/£)    be   found   for   any 
assumed  axis  through  the  given  center  by  one  of  the  methods 
already  described  (Arts.  202  and   203).     Then   two  lines   par- 
allel to  the  axis  and  distant  k  from  it,  on  opposite  sides,  will  be 
tangents  to  the  inertia-ellipse. 

213.  To  Construct  the  Inertia-Ellipse,  a  Pair  of  Conjugate 
Axes  Being  Known  in  Position.  —  If  the  positions  of  two  axes 
conjugate  to  each  other  can  be  found,  the  ellipse  can  be  drawn 
by  the  following  method  : 

Determine  the  radius  of  gyration  for  each  of  the  two  axes 
and  draw  the  corresponding  tangents  as  in  the  preceding 
article;  then  proceed  as  follows: 

Let  XX',  YY'  (Fig.  73)  be  the  given  axes,  and  let  the  four 

tangents  determined  as  above  form  the  parallelogram  PQRS. 

•     /Y  Let  A,  A',  B,  B'  be  the  points 

P  _  BZ.  -  -,  Q      in    which    these    tangents    in- 

/  /  /          tersect   the   axes   XX',    YY'. 


o  /A  Then,  since  each  diameter  is 

/  parallel  to  the  tangents  at  the 


S          B/  R  extremities   of   the    conjugate 

/Y'  Fig. 73  diameter,  A,  A',  B,  B'  are  the 

extremities    of   the   diameters   lying  in  the  given   axes.     The 

ellipse    can    now    be    constructed    as    explained   in   Art.    210 

(Fig,  72). 

This    method  of    constructing    the    inertia-ellipse   is    useful 


INERTIA-CURVES   FOR   PLANE   AREAS.  187 

whenever  the  given  system  has  a  pair  of  conjugate  axes  which 
can  be  located  by  inspection. 

214.  Central  Ellipse. — It   is   evident   that    an   inertia-curve 
can   be  found  with   its    center   at  ,any  assumed    point.     That 
ellipse  whose  center  is  the  centroid  of  the  given  system  is  called 
the  central  ellipse  for  the  system. 

Since  the  central  ellipse  gives  at  once  the  radius  of  gyration 
for  every  axis  through  the  centroid  of  the  system,  it  enables  us 
to  determine  readily  the  radius  of  gyration  for  any  axis  what- 
ever, by  means  of  the  known  relation  between  radii  of  gyration 
for  parallel  axes.  (Art.  196.) 

§  3.    Inertia-Curves  for  Plane  Areas. 

215.  General  Principles.  —  The    principles   deduced   in   the 
treatment  of  inertia-curves  for  systems  of  forces  are  all  true  for 
the  case  of  plane  areas.     But  special  difficulties  arise  in  dealing 
with  areas,  because  of  the  fact  that  the  system  of  forces  equiva- 
lent to  any  area  consists  of  an  infinite  number  of  forces.     The 
principles  already  developed  are,  however,  sufficient  to  deal  at 
least  approximately  with  all  areas,  and   accurately  with  many 
cases. 

216.  Inertia-Curve  an  Ellipse.  —  Since  the  forces  conceived 
to  replace  the   elements  of  area  (Art.    198)  have  all  the  same 
sign,  the  value  of  ft  is  always  positive,  and  the  inertia-curve  is 
always  an  ellipse.     (Arts.  209  and  211.) 

217.  Cases  Admitting  Simple  Treatment.  —  Whenever  a  pair 
of  conjugate  diameters  can  be  located,  and  the  radius  of  gyra- 
tion determined   for  each,  the  inertia-ellipse   can    be   at    once 
drawn  as  in  Art.  213.     This  will  be  the  case  whenever  it  is 
possible  to  locate  readily  a  pair  of  axes  for  which  the  product  of 
inertia  is  zero. 

(i)  If  there  is  an  axis  of  symmetry,  this  and  any  line  perpen- 
dicular to  it  are  a  pair  of  conjugate  axes  (and  in  fact  the  principal 
axes)  of  the  inertia-ellipse  whose  center  is  at  their  intersection. 
(Art.  200.) 


1 88  GRAPHIC   STATICS. 

(2)  If  two  axes  can  be  located  in  such  a  way  that  for  every 
element  of  area  whose  distances  from  the  axes  are  /,  q,  there  is 
an  equal  element  whose  distances  are  /,  —q,  or  —  p,  g,  the 
product  of  inertia  is  zero  for  the  two  axes,  and  these  are  there- 
fore a  pair  of  conjugate  axes  of  the  inertia-ellipse  whose  center 
is  at  their  intersection.  This  of  course  includes  the  case  when 
there  is  an  axis  of  symmetry.  (Art.  200.) 

When  a  pair  of  conjugate  axes  is  known,  the  radius  of  gyration 
is  to  be  found  by  one  of  the  methods  of  Art.  202  or  Art.  203  ; 
the  ellipse  can  then  be  drawn  exactly  as  explained  in  Art.  213. 

If  a  pair  of  conjugate  axes  cannot  be  located  by  inspection, 
the  inertia-ellipse  cannot  be  so  readily  constructed.  Such  cases 
will  not  be  here  treated. 

As  examples  of  areas,  in  which  the  principal  axes  of  the 
inertia-curve  can  be  located  by  inspection,  may  be  mentioned 
the  cross-section  of  the  I-beam,  the  deck-beam,  the  channel-bar, 
and  other  .shapes  of  structural  iron  and  steel. 

Many  geometrical  figures  possess  axes  of  symmetry.  In 
others  a  pair  of  conjugate  axes  can  be  located  by  principle  (2). 
Some  of  these  will  be  discussed  in  the  next  article. 

Example,  —  Draw  the  central  ellipse  for  the  deck-beam  sec- 
tion shown  in  Fig.  74. 

[SUGGESTIONS.  —  Since  there  is  an  axis  of  symmetry,  this 
contains  one  of  the  principal  axes  of  the  ellipse.  The  other 
can  be  drawn  as  soon  as  the  centroid'  of  the  section  is 
known.  Find  the  radius  of  gyration  for  each  axis  by 
Art.  202,  and  then  construct  the  ellipse  as  explained  in 
Art.  213.] 

2 1 8.    Central  Ellipses  for  Geometrical  Fig- 
ures. —  In  many  of  the  simple  geometrical 
figures,   not    only  can   a   pair  of   conjugate 
axes  be  located  by  inspection,  but  the  radius 
of  gyration  for  each  of  these  axes  can  be  found  by  a  simple 
construction,  so  that  the  central  ellipse  can  be  readily  drawn. 
Some  of  these  cases  will  be  here  summarized. 


INERTIA-CURVES    FOR   PLANE   AREAS.  189 

(i)  Parallelogram.—  Let  ABCD  (Fig.  75)  be  the  parallelo- 
gram; then  XX',  YY',  drawn  through  the  centroid  parallel  to 
the  sides,  are  a  pair  of 
conjugate  axes  of  the  cen- 
tral ellipse.  Let  AB  =  b, 
BC=d,  and  let  //=  the 
perpendicular  distance  be- 
tween AB  and  DC.  The 
moment  of  inertia  of  the 
parallelogram  with  re- 
spect to  the  axis  XX1  is 
equal  to  the  moment  of  inertia  of  a  rectangle  of  sides  b  and  //. 
Hence  £2,  the  square  of  the  radius  of  gyration  for  this  axis,  is 

/2 

—  •     The  length  of  k  can  be  found  by  the  construction  used  in 

case  of  the  rectangles  in  Fig.  70.     The  following  modification 
of  the  method  is,  however,  more  convenient  : 

Make  EF=\BC,  EG  =  \BC,  and  draw  a  semicircle  with  FG 
as  a  diameter.  From  E  draw  a  line  perpendicular  to  BC,  inter- 
secting the  semicircle  at  /.  Lay  off  EH=EI;  then  a  line 
through  H  parallel  to  XX*  is  a  tangent  to  the  central  ellipse. 
For  by  construction, 


. 

Vl2 

And  since  the  projection  of  BC  on  a  line  perpendicular  to  XX  is 
equal  to  h,  the  projection  of  EH  on  the  same  line  is  equal  to  —  l--, 

Vl2 

that  is  to  k. 

The  tangent  parallel  to  the  side  BC  may  be  found  in  a  simi- 
lar way.  It  may,  however,  be  located  more  simply  as  follows  : 
It  is  evident  that  the  distance  between  YY'  and  a  tangent  par- 
allel to  it,  measured  along  AB,  bears  the  same  ratio  to  AB  that 
EH  does  to  BC.  Hence,  the  parallelogram  formed  by  the  four 
tangents,  two  parallel  to  XX'  and  two  parallel  to  YY1,  is  simi- 
lar to  the  parallelogram  ABCD. 


1 90 


GRAPHIC    STATICS. 


Fig.  75  shows  this  parallelogram  and  also  the  ellipse. 

(2)  Triangle.— Let   ABC   (Fig.  76)  be   the    triangle;   b  = 

length  of  the  base  BC\  b'  = 
length  of  projection  of  BC  on  a 
line  perpendicular  to  AD ;  d  =  al- 
titude measured  perpendicular  to 
BC.  AD  and  a  line  through  the 
centroid  parallel  to  BC  are  a  pair 
of  conjugate  axes  of  the  central 
ellipse  (Art.  217).  From  Art. 
199,  the  radius  of  gyration  for  a 
central  axis  parallel  to  BC  is 

Let  XX'  be  this  axis,  and  H  the  point  in  which 
18         30 

it  intersects  AC.  Then  HC=%AC.  Take  HK=\AC=\HC\ 
and  make  KC  the  diameter  of  a  semicircle.  From  H  draw  HI 
perpendicular  to  AC,  intersecting  the  semicircle  at  /.  Make 
HL  =  HI\  then  the  line  through  L  parallel  to  XX'  is  a  tangent 
to  the  central  ellipse.  For  the  radius  of  gyration  with,  respect 
to  XX1  is  to  HL  as  the  altitude  ^is  to  AC. 

Again,  for  the  axis  AD,  the  radius  of  gyration  is  -y —  (Art. 

199).  Make  DE=\  BC  and  DF=\  BC,  and  take  EF  as  a 
diameter  of  a  semicircle.  From  D  draw  a  line  perpendicular  to 
BC,  intersecting  the  semicircle  in  M\  and  make  DG=DM\  then 
a  line  from  G  parallel  to  AD  is  a  tangent  to  the  central  ellipse. 
The  figure  shows  the  parallelogram  formed  by  the  two  tan- 
gents parallel  to  XX'  and  the  two  parallel  to  AD,  and  also  the 
central  ellipse. 

(3)  Ellipse.  —  From   Art.    199,  the   radii  of   gyration  of   an 
ellipse   with    respect    to  the   two    principal    diameters  are   \a 
and  -|-  b.     Hence  the  central  ellipse  of  inertia  is  similar  to  the 
given  ellipse^  its  semi-axes  being  -|  a  and  |  b.     A  special  case 
of  this  is  a  circle,  for  which  the  central  curve  is  a  circle  whose 
radius  is  half  that  of  the  given  circle. 


INERTIA-CURVES    FOR   PLANE   AREAS. 


191 


(4)  Semicircle.  —  Let  ABC  (Fig.  77)  be  the  semicircle,  O 
being  the  centroid.  (The  point  O  may  be  located  by  the 
method  described  in  Art.  184.) 
From  symmetry  it  is  evident 
that  the  principal  axes  of  the 
central  ellipse  are  XX'  and 
YY',  drawn  through  0,  re- 
spectively parallel  and  per- 
pendicular to  AB. 

With  respect  to  the  axis 
YY1,  the  radius  of  gyration 
is  evidently  £  r,  the  same  as 

for  the  whole  circle.     Hence  two  lines  parallel  to  YY'  and  dis- 
tant J  r  from  it  are  tangents  to  the  central  ellipse. 

Again,  the  radius  of  gyration  of  the  semicircle  with  respect 
to  AB  as  an  axis  is  also  \  r,  the  same  as  for  the  whole  circle. 
To  find  it  for  the  axis  XX',  with  D  as  a  center,  and  radius  £  r, 
draw  an  arc  intersecting  XX'  at  F\  then  ^F2  =  DF2~aff. 
But  DF  is  equal  to  the  radius  of  gyration  with  respect  to  AB, 
and  <9Z>is  the  distance  between  XX1  and  AB ';  hence  (Art.  196) 
OF  is  equal  to  the  radius  of  gyration  with  respect  to  XX'. 
Hence  if  two  lines  are  drawn  parallel  to  XX',  each  at  a  distance 
from  it  equal  to  OF,  they  will  be  tangents  to  the  central  ellipse. 
The  ellipse  can  now  be  drawn  in  the  usual  manner. 

219.  Summary  of  Results.  —  By  the  principles  and  methods 
developed  in  the  present  chapter,  inertia-curves  can  be  drawn 
for  all  the  simpler  cases  that  may  arise ;  namely,  whenever  a 
pair  of  conjugate  axes  can  be  located  by  inspection.  This  will 
be  the  case  whenever  the  product  of  inertia  can  be  seen  to  be 
zero  for  any  pair  of  axes  ;  and  it  includes  every  case  of  an  area 
possessing  an  axis  of  symmetry. 

It  is  believed  that  this  chapter  contains  as  complete  a  discus- 
sion as  is  needed  by  the  student  of  engineering.  Those  who 
desire  to  pursue  the  subject  further  may  consult  other  works. 


.  34 


U) 

Scale,  1  inch  =  6  feet 


eft  .     Scale,  linch= GOO  Ibs. 

\ 


PLATE  I. 


DEF 
Scale,  1  inch=6,ooo  Ibs. 

(C) 


Scale:  I  in.=  4.OOO  Ibs. 


Scale:  I  in, 


Fig-.  42 


PLATE    II. 


1 

, 

1 

1             1 

| 

1 

1 

oo 

8 

3           S" 

9 

cT 

» 

Si 

8.1 

5.8       1      4.5 

4.5 

7.1 

4.8 

5.7 

t 

°      OOOOooo 


Linear  Scale,  1  in.  =  lo  ft. 
Force  Scale,  1  in.  =  80,000  Ibs. 


Xmr 


Xs 


xt 


PLATE  III. 


8.1  5.8 


7.1  4.8 


o      OOOO     oooo 


a      a     3     a 

8.1  5.8         4.5       4.5  7.1       I    4.8    I     5.7 


o    O  OOO  oo  oc 


Fig.oO 

Linear  Scale,  1  in.=SOft. 
Force  Scale,  1  in. =80,000  Ibs. 


PLATE  IV. 


a      a      a 

5.8     I    4.5    I    4.5    I       7.1       I    4.8    I     o.7     I    4.8 


f,.0 


ill 

I  4.0  I  4.0 


o     O  OOO    O  O  O  Q  i  »*•"-»"•• 


S  9  T  10  M  11  V 


Linear  Scale,  1  in.-'. 
Force  Scale,  1  in.— 80,000  Ibs. 


PLATE  V. 


8      S 


4.7    U.O  9.0 


7.4  7.3          7.8 


1.0 


11 

7    U.O  I 


oo  oo     o    O  O  O   oo  oo 


X  I          P          j  Q          k          R 


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Solutions  of  the  Examples  contained  in  the  Above.     In  the  Press. 

THE  ELEMENTS    OF   STATICS   AND   DYNAMICS. 

Parti.  Elements  of  Statics.  $1.25.  Part  II.  Elements  of  Dynamics. 
$1.00.  Complete  in  one  volume.  i6mo.  $1.90. 

AN  ELEMENTARY   TREATISE   ON   KINEMATICS   AND   DYNAMICS. 

By  JAMES  GORDON  MACGREGOR,  M.A.,  D.Sc.,  Munro  Professor  of  Physics, 
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The  problems  and  exercises  seem  to  be  well  chosen,  and  the  book  is  provided 
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WORKS  BY  G.  M.  MINCHIN,  M.A., 

Professor  of  Mathematics  in  the  Royal  Indian  Engineering  College. 

A  TREATISE  ON   STATICS. 

Third  Edition,  corrected  and  enlarged.  Vol.  I.  Equilibrium  of  Coplanar 
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UNIPLANAR   KINEMATICS   OF   SOLIDS   AND   FLUIDS.     i2mo.    $1.90. 

A  TREATISE   ON  ELEMENTARY  MECHANICS. 

For  the  use  of  the  Junior  Classes  at  the  Universities,  and  the  Higher 
Classes  in  Schools.  With  a  collection  of  examples  by  S.  PARKINSON, 
F.R.S.  Sixth  Edition.  i2mo.  $2.25. 

LESSONS   ON   RIGID   DYNAMICS. 
By  the  Rev.  G.  PIRIE,  M.A.     121110.     $1.50. 

ELEMENTARY   STATICS. 

By  G.  RAWLINSON,  M.A.     Edited  by  E.  STURGES.     8vo.     $1.10. 

WORKS  BY  E.  J.  ROUTH,  LL.D.,  F.R.S., 

Fellow  of  the  Senate  of  the  University  of  London. 

A  TREATISE  ON  THE  DYNAMICS  OF  A  SYSTEM  OF  RIGID  BODIES. 
With  Examples.     New  Edition  Revised  and  Enlarged.     8vo. 

Part  I.  Elementary.     Fifth  Edition  Revised  and  Enlarged.     $3.75. 
Part  II.  Advanced.     $3.75. 


4  MECHANICS.  — HIGHER  PURE  MATHEMATICS. 

STABILITY   OF  A  GIVEN   STATE   OF  MOTION, 

Particularly  Steady  Motion.     8vo.     $2.25. 

HYDROSTATICS   FOR  BEGINNERS. 
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Mr.  Sanderson's  work  has  many  merits,  and  is  evidently  a  valuable  text-book, 
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SYLLABUS   OF  ELEMENTARY   DYNAMICS. 

Part  I.  Linear  Dynamics.  With  an  Appendix  on  the  Meanings  of  the 
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A  TREATISE  ON   DYNAMICS   OF  A  PARTICLE. 

By  P.  G.  TAIT,  M.A.,  and  W.  J.  STEELE.     Sixth  Edition,  Revised.     $3.00. 

WORKS  BY  ISAAC  TODHUNTER,  F.R.S. 

Late  Fellow  and  Principal  Mathematical  Lecturer  of  St.  Johns  College. 

MECHANICS    FOR   BEGINNERS. 
With  Numerous  Examples.    New  Edition.     i8mo.    $1.10.    KEY.  #1.75. 

A  TREATISE  ON  ANALYTICAL   STATICS. 

Fifth  Edition.     Edited  by  Prof.  J.  D.  EVERETT,  F.R.S.     i2mo.     $2.60. 

A   COLLECTION   OF  PROBLEMS   IN  ELEMENTARY  MECHANICS. 
By  W.  WALTON,  M.A.     Second  Edition.     $1.50. 

PROBLEMS   IN   THEORETICAL  MECHANICS. 

Third  Edition,  Revised.  With  the  addition  of  many  fresh  Problems. 
By  W.  WALTON,  M.A.  8vo.  $4.00. 

ttriQber  pure  flftatbematics, 

WORKS  BY  SIR  G.  B.  AIRY,  K.C.B.,  formerly  Astronomer-Royal. 
ELEMENTARY  TREATISE  ON  PARTIAL  DIFFERENTIAL  EQUATIONS. 
With  Diagrams.     Second  Edition.     12 mo.     $1.50. 

ON  THE  ALGEBRAICAL  AND  NUMERICAL  THEORY  OF  ERRORS 

of  Observations  and  the  Combination  of  Observations.  Second  Edition, 
revised.  121110.  $1.75. 

NOTES   ON   ROULETTES   AND   GLISSETTES. 

By  W.  H.  BESANT,  D.Sc.,  F.R.S.     Second  Edition,  Enlarged.     $1.25. 
A  TREATISE   ON   THE   CALCULUS   OF  FINITE   DIFFERENCES. 
By  the  late  GEORGE  BOOLE.     Edited  by  J.  F.  MOULTON.     Third  Edition. 
i2mo.     $2.60. 

ELEMENTARY  TREATISE   ON  ELLIPTIC   FUNCTIONS. 

By  ARTHUR  CAYLEY,  D.Sc.,  F.R.S.     New  Edition  preparing. 


HIGHER   PURE  MATHEMATICS. 


DIFFERENTIAL   CALCULUS. 

With  Applications  and  Numerous  Examples.  An  Elementary  Treatise. 
By  JOSEPH  EDWARDS,  M. A.  i2mo.  $2.75. 

A  useful  text-book  both  on  account  of  the  arrangement,  the  illustrations,  and  ap- 
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given.  —  Educational  Times. 

AN  ELEMENTARY   TREATISE   ON   SPHERICAL  HARMONICS 

And  Subjects  Connected  with  Them.  By  Rev.  N.  M.  FERRERS,  D.D., 
F.R.S.  121110.  $1.90. 

WORKS  BY  ANDREW  RUSSELL  FORSYTH,  M.A. 
A   TREATISE   ON   DIFFERENTIAL  EQUATIONS.     8vo.     $3.75. 
As  Boole's  "  Treatise  on  Different  Equations  "  was  to  its  predecessor,  Wymer's, 
this  treatise  is  to  Boole's.     It  is  more  methodical  and  more  complete,  as  far  as  it 
goes,  than  any  that  have  preceded  it.  —  Educational  Times. 

THEORY  OF  DIFFERENTIAL   EQUATIONS. 

Part  I.    Exact  Equations  and  Pfaff's  Problems.     8vo.     $3.75. 

AN   ELEMENTARY   TREATISE   ON   CURVE   TRACING. 

By  PERCIVAL  FROST,  M.A.     8vo.     $3.00. 

DIFFERENTIAL  AND  INTEGRAL   CALCULUS. 

With  Applications.    By  ALFRED  GEORGE  GREENHILL,  M.A.    i2mo.    $2.00. 

APPLICATION   OF  ELLIPTIC   FUNCTIONS. 

By  ALFRED  GEORGE  GREENHILL,  M.A.     i2mo.     $3.50. 

AN   ELEMENTARY  TREATISE   ON   THE   DIFFERENTIAL  AND 

INTEGRAL  CALCULUS. 
By  G.  W.  HEMMING,  M.A.     8vo.     $2.50. 

THEORY   OF   FUNCTIONS. 

By  JAMES  HARKNESS,  M.A.,  Professor  of  Mathematics,  Bryn  Mawr  Col- 
lege, and  FRANK  MORLEY,  A.M.,  Professor  of  Mathematics,  Haver- 
ford  College,  Pa.  In  preparation. 

INTRODUCTION   TO   QUATERNIONS. 

With  Examples.  By  P.  KELLAND,  M.A.,  and  P.  G.  TAIT,  M.A.  Second 
Edition.  i2mo.  $2.00. 

HOW   TO   DRAW   A   STRAIGHT   LINE. 

A  Lecture  on  Linkages.  By  A.  B.  KEMPE,  B.A.  With  Illustrations. 
Nature  Series.  12 mo.  50  cents. 

DIFFERENTIAL   CALCULUS   FOR   BEGINNERS. 

With  Examples.     By  ALEXANDER  KNOX,  B.A.     i6mo.     90  cents. 

MESSENGER    OF    MATHEMATICS. 

Edited  by  J.  W.  L.  GLAISHER.    Published  Monthly.    35  cents  each  number. 

WORKS  BY  THOMAS  MUIR, 

Mathematical  Master  in  the  High  School,  Glasgow. 
A  TREATISE   ON  THE   THEORY  OF  DETERMINANTS. 
With  Examples.     12 mo.     New  Edition  in  Preparation. 


HIGHER   PURE  MATHEMATICS. 


THE   THEORY   OF   DETERMINANTS 

In  the  Historical  Order  of  its  Development.  Part  I.  Determinants  in 
General.  Leibnitz  (1693)  to  Cayley  (1841).  8vo.  $2.50. 

TREATISE   ON   INFINITESIMAL   CALCULUS. 
By  BARTHOLOMEW  PRICE,  M.A.,  F.R.S.,  Professor  of  Natural  Philosophy, 

Oxford. 

Vol.  I.    Differential  Calculus.     Second  Edition.     8vo.     $3.75. 

Vol.  II.    Integral  Calculus,  Calculus  of  Variations,  and  Differential 
Equations.     8vo.     Reprinting. 

Vol.  III.    Statics,  including  Attractions  ;  Dynamics  of  a  Material  Par- 
ticle.    8vo.     $4.00. 

Vol.  IV.    Dynamics  of  Material  Systems.     Together  with  a  Chapter 
on  Theoretical  Dynamics,  by  W.  F.  DONKIN,  M.A.     8vo.     $4.50. 

A   TREATISE   ON   THE   THEORY  OF  DETERMINANTS, 
And  their  Applications  in  Analysis  and  Geometry.     By  ROBERT  SCOTT, 
M.A.     8vo.     $3.50. 

FACTS   AND   FORMULAE   IN   PURE   MATHEMATICS 

And  Natural  Philosophy.  Containing  Facts,  Formulae,  Symbols,  and 
Definitions.  By  the  late  G.  R.  SMALLEY,  B.A.,  F.R.A.S.  New  Edi- 
tion by  J.  M'DOWELL,  M.A.,  F.R.A.S.  i6mo.  70  cents. 

MATHEMATICAL  PAPERS  OF  THE  LATE  REV.  J.  S.  SMITH. 

Savilian  Professor  of  Geometry  in  the  University  of  Oxford.  u  With 
Portrait  and  Memoir.  2  vols.  4to.  In  preparation. 

AN  ELEMENTARY  TREATISE  ON  QUATERNIONS. 

By  P.  G.  TAIT,  M.A.,  Professor  of  Natural  Philosophy  in  the  University 
of  Edinburgh.  Third  Edition,  Much  Enlarged.  8vo.  $5.50. 

WORKS  BY  ISAAC  TODHUNTER,  F.R.S., 

Late  Principal  Lecturer  on  Mathematics  in  St.  Johns  College,  Cambridge. 
AN  ELEMENTARY  TREATISE  ON  THE  THEORY  OF  EQUATIONS. 
i2mo.     $1.80. 

A   TREATISE   ON   THE   DIFFERENTIAL   CALCULUS. 

121110.     $2.60.     KEY.     $2.60. 

A   TREATISE   ON   THE   INTEGRAL   CALCULUS 

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AN   ELEMENTARY   TREATISE   ON   LAPLACE'S, 

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TRILINEAR   CO-ORDINATES, 

And  Other  Methods  of  Modern  Analytical  Geometry  of  Two  Dimen- 
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8vo.  $4.00. 


ANAL  YTICAL    GE OME  TR  Y. 


analytical  geometry    (PLANE  AND  SOLID 

SOLID  GEOMETRY. 

By  W.  STEADMAN  ALDIS.     121110.     $1.50. 

ELEMENTS  OF  PROJECTIVE   GEOMETRY. 

By  LUIGI  CREMONA,  LL.D.  Translated  by  CHARLES  LEUDESDORF,  M.A. 
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EXERCISES  IN  ANALYTICAL  GEOMETRY. 
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A  TREATISE  ON  TRILINEAR  CO-ORDINATES, 
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the  Rev.  N.  M.  FERRERS.     Fourth  Edition.     i2mo.     $1.75. 

WORKS  BY  PERCIVAL  FROST,  D.Sc.,  F.R.S. 

Lecturer  in  Mathematics,  King's  College,  Cambridge. 

AN  ELEMENTARY  TREATISE  ON  CURVE  TRACING.   8vo.   $3.00. 
SOLID  GEOMETRY. 

A  New  Edition,  revised  and  enlarged,  of  the  Treatise,  by  FROST  and 
WOLSTENHOLME.  Third  Edition.  8vo.  $6.00. 

Hints  for  the  Solution  of  Problems  in  the  above.    8vo.     $3.00. 

THE   ELEMENTS  OF  SOLID  GEOMETRY. 
By  R.  BALDWIN  HAYWARD,  M.A.,  F.R.S.     i2mo.     75  cents. 
THE  GEOMETRY  OF  THE   CIRCLE. 

By  W.  J.  MCCLELLAND,  M.A.,  Trinity  College,  Dublin  ;  Head  Master  of 
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AN  ELEMENTARY  TREATISE  ON  CONIC  SECTIONS 

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By  G.  HALE  PUCKLE,  M.A.  Fifth  Edition,  Revised  and  Enlarged. 
i2mo.  $1.90. 

WORKS  BY  CHARLES  SMITH,  M.A. 

Fellow  and  T^ltor  of  Sidney  Sussex  College,  Cambridge. 
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Seventh  Edition.     i2mo.     $1.60.     KEY.     $2.60. 

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AN  ELEMENTARY  TREATISE  ON  SOLID  GEOMETRY.    i2mo.   $2.60. 
WORKS  BY  ISAAC  TODHUNTER,  F.R.S. 

Late  Principal  Lecturer  on  Mathematics  in  St.  John's  College. 

PLANE  CO-ORDINATE   GEOMETRY, 

As  Applied  to  the  Straight  Line  and  the  Conic  Sections.  i2mo.  $1.80. 
KEY.  By  C.  W.  BOURNE,  M.A.  i2mo.  $2.60. 

EXAMPLES  IN  ANALYTICAL  GEOMETRY 

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ANALYTICAL  GEOMETRY  FOR  SCHOOLS. 
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A  TREATISE   ON  ALGEBRA. 

By  Charles  Smith,  M.A. 

REVISED  AND   ENLARGED   EDITION,  $1.90. 

„,%  This  new  edition  has  been  greatly  improved  by  the  addition  of  a  cJiapter 
on  Differential  Equations  and  other  changes. 


u  Your  Smith's  '  Treatise  on  Algebra '  was  used  in  our  University  classes 
last  session,  and  with  very  great  satisfaction.   .   .   .     The  general  adoption  . 
of  these  texts  would  mark  an  epoch  in  mathematical  teaching."  —  Prof.  W.  B. 
SMITH,  University  of  Missouri. 

"  Those  acquainted  with  Mr.  Smith's  text-books  on  conic  sections  and 
solid  geometry  will  form  a  high  expectation  of  this  work,  and  we  do  not  think 
they  will  be  disappointed.  Its  style  is  clear  and  neat,  it  gives  alternative 
proofs  of  most  of  the  fundamental  theorems,  and  abounds  in  practical  hints, 
among  which  we  may  notice  those  on  the  resolution  of  expression  into  factors 
and  the  recognition  of  a  series  as  a  binomial  expansion."' —  Oxford  Review. 

HIGHER  ALGEBRA. 

'By  H.  S.  Hall,  M.A.,  and  S.  R.  Knight,  B.A. 

A  Sequel  to  the  Elementary  Algebra  for  Schools  by  the  same 
Authors.  Fourth  edition,  containing  a  collection  of  three 
hundred  Miscellaneous  Examples  which  will  be  found  useful 
for  advanced  students.  I2mo.  $1.90. 

** .  .  .  It  is  admirably  adapted  for  College  students  as  its  predecessor 
was  for  schools.  It  is  a  well  arranged  and  well  reasoned-out  treatise,  and 
contains  much  that  we  have  not  met  with  before  in  similar  works.  For 
instance,  we  note  as  specially  good  the  articles  on  Convergency  and  Diver- 
gency of  Series,  on  the  treatment  of  Series  generally,  and  the  treatment  of 
Continued  Fractions.  .  .  .  The  book  is  almost  indispensable  and  will  be 
found  to  improve  upon  acquaintance." —  The  Academy. 

•'  We  have  no  hesitation  in  saying  that,  in  our  opinion,  it  is  one  of  the 
best  books  that  have  been  published  on  the  subject.  .  .  .  The  last  chap- 
ter supplies  a  most  excellent  introduction  to  the  Theory  of  Equations.  We 
would  also  specially  mention  the  chapter  on  Determinants  and  their  appli- 
cation, forming  a  useful  preparation  for  the  reading  of  some  separate  work 
on  the  subject.  The  authors  have  certainly  added  to  their  already  high  repu- 
tation as  writers  of  mathematical  text-books  by  the  work  now  under  notice, 
which  is  remarkable  for  clearness,  accuracy,  and  thoroughness.  .  .  .  Al- 
though we  have  referred  to  it  on  many  points,  in  no  single  instance  have 
we  found  it  wanting." —  The  School  Guardian. 


MACMILLAN    &   CO.,   NEW  YORK. 


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